ap bio midterm

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Three nucleotides code for: A. 1 amino acid B. 3 amino acids C. 1 protein D. 3 proteins

A. 1 amino acid

In RNA, _____ codon(s) translate to ______ amino acid(s) A. 1, 1 B. 3, 1 C. 3, 3 D. 1, 20

A. 1,1

If a DNA molecule is found to be composed of 40% thymine, what percentage of guanine would be expected. A. 10% B. 20% C. 40% D. 80%

A. 10%

The diploid cells of an animal have 44 chromosomes. How many chromosomes would you expect to find in the animal's sperm cells? A. 22 B. 44 C. 88 D. 2

A. 22

After examining a karyotype of an animal, a researcher counts 50 individual chromosomes of varying sizes. How many homologs are in the karyotype? A. 25 B. 50 C. 100 D. 200

A. 25

Researchers have discovered details about apoptosis (programmed cell death) by studying embryologic development of a nematode worm, Caenorhabditis elegans. Apoptosis is a normal developmental process in C. elegans. They found several genes involved in apoptosis, including ced−9 and ced−3 . The ced−3 gene was found to promote cell death, and ced−9 to inhibit it. The ced−9 gene serves as a regulator that prevents apoptosis in the absence of a signal promoting apoptosis. Which of the following statements best justifies the claim that changes in the expression of ced−9 in C. elegans can affect regulation of apoptosis in the cell? A. An experiment showed that a mutation in the ced−9 gene led to excessive cell death in C. elegans. B. An experiment showed that the ced−9 gene normally produces a protein that promotes excessive cell death in C. elegans. C. A mutation in ced−3 will cause ced−9 to be incorrectly transcribed. D. Apoptosis is dependent on a signal from the ced−9 gene in C. elegans.

A. An experiment showed that a mutation in the ced−9 gene led to excessive cell death in C. elegans.

In vascular plants, water flows from root to leaf via specialized cells called xylem. Xylem cells are hollow cells stacked together like a straw. A student explains that evaporation of water from the leaf pulls water up from the roots through the xylem, as shown in Figure 1. Figure 1. Model of water movement through the xylem, with magnified models of water movement in the stem and leaf. Which statement describes how water is pulled up through the xylem to the leaves of the plant? A. As water exits the leaf, hydrogen bonding between water molecules pulls more water up from below. B. As water exits the leaf, signals are sent to the roots to pump more water up to the leaves through the xylem by adhesion. C. Evaporation from the leaf decreases the hydrogen bonds that form between the water molecules in the xylem, which helps the water molecules to be pulled up the xylem. D. Evaporation of water from the leaf increases the hydrogen bonds that form between water molecules in the air, providing the energy for transport.

A. As water exits the leaf, hydrogen bonding between water molecules pulls more water up from below.

Several members of a family are found to involuntarily sneeze when exposed to bright lights, such as sunlight. Following analysis of the condition in the family, a doctor predicts that the symptoms have an underlying genetic basis. A pedigree of the family is shown in Figure 1. Figure 1. Pedigree showing family members with and without symptoms For this condition, which of the following modes of inheritance is most consistent with the observations? A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. X-linked recessive

A. Autosomal dominant

During a fight-or-flight response, epinephrine is released into the body's circulatory system and transported throughout the body. Some cells exhibit a response to the epinephrine while other cells do not. Which of the following justifies the claim that differences in components of cell signaling pathways explain the different responses to epinephrine? A. Cell signaling depends on the ability to detect a signal molecule. Not all cells have receptors for epinephrine. Only cells with such receptors are capable of responding. B. Cell signaling depends on the transduction of a received signal by the nervous system. Not all cells are close enough to a synapse to receive the signal and respond. C. Cell signaling depends on the signal being able to diffuse through the cell membrane. Epinephrine is incapable of diffusing through some plasma membranes because of the membrane's phospholipid composition. D. Cell signaling requires reception, transduction, and response. All cells can receive epinephrine, all cells respond with a pathway, but only select cells have the proper coding in their DNADNA to respond.

A. Cell signaling depends on the ability to detect a signal molecule. Not all cells have receptors for epinephrine. Only cells with such receptors are capable of responding.

An antigen can induce an immune response in a host organism. Antigens are targeted by antibodies that are produced by the organism's immune system in response to contact with the antigen. Antibodies are specific to antigens. Many different cell types are involved in immune responses. Which of the following best describes the role of cell-to-cell communication during a response to an invasion of antigens? A. Chemicals that are secreted from antigen-presenting cells then activate helper T cells. B. A macrophage cell engulfs a pathogen in the blood. C. Antigens attaching to receptors on memory B cells stimulate the memory B cells to become plasma cells. D. Antigen-presenting cells engulf antigens at the first exposure.

A. Chemicals that are secreted from antigen-presenting cells then activate helper T cells.

What was the overall conclusion of the hershey-chase experiment A. DNA was responsible for heredity B. proteins and DNA were responsible for heredity C. the ratio of Adenine to thymine was always the same D. phage DNA was similar to bacterial DNA

A. DNA was responsible for heredity

A model of crossing over during gamete formation is shown in Figure 1. Figure 1. Model of crossing over during meiosis Based on Figure 1, which of the following questions could best be addressed? A. Does synapsis of homologous chromosomes in the parent cell contribute to an increase in genetic diversity in the daughter cells? B. Do sister chromatids separate and form diploid daughter cells? C. Do chromatids from nonhomologous chromosomes rearrange to produce identical daughter cells? D. Does synapsis of homologous chromosomes produce daughter cells that are identical to the parent cell?

A. Does synapsis of homologous chromosomes in the parent cell contribute to an increase in genetic diversity in the daughter cells?

Figure 1 shows three amino acids that are part of a polypeptide chain. Figure 2 shows the same section of the chain after a mutation has occurred. How might this change affect the structure and function of the protein? A. The R-group of the new amino acid, valine, has different chemical properties than the R-group of cysteine. This will cause the protein to misfold and not function properly in the cell. B. The new amino acid, valine, has replaced cysteine in the new protein. Since the number of amino acids has remained the same, there will be no change in the three-dimensional folding, or function, of the protein. C. Since this is a linear section, it does not influence protein folding. Thus, there will be no change in protein structure or function. D. Since the new amino acid is bounded on one side by an amino acid with a negatively charged R-group and by an amino acid on the other side with a positively charged R-group, the charges will balance and the protein will fold as usual.

A. The R-group of the new amino acid, valine, has different chemical properties than the R-group of cysteine. This will cause the protein to misfold and not function properly in the cell.

Researchers hypothesized that red eye color in Japanese koi, a type of fish, is due to a mutation. To study the inheritance of red eye color in koi, the researchers conducted breeding experiments over several generations. The results are summarized in Figure 1. Figure 1. Inheritance of eye color in Japanese koi Based on the data in Figure 1, which of the following is the best prediction of the mode of inheritance of red eyes in Japanese koi? A. The allele for red eyes is inherited in an autosomal dominant pattern. B. The allele for red eyes is inherited in an autosomal recessive pattern. C. The allele for red eyes is inherited in an X-linked recessive pattern. D. The allele for red eyes is inherited in an X-linked dominant pattern.

A. The allele for red eyes is inherited in an autosomal dominant pattern.

Eukaryotes transcribe RNA from DNA that contains introns and exons. Alternative splicing is one posttranscriptional modification that can create distinct mature mRNA molecules that lead to the production of different proteins from the same gene. Figure 1 shows a gene and the RNA produced after transcription and after alternative splicing. Figure 1. Model of posttranscriptional alternate splicing of mRNA A cell needs to metabolize the substrate illustrated in Figure 1 for a vital cellular function. Which of the following best explains the long-term effect on the cell of splicing that yields only enzyme C mRNA? A. The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown. B. The cell will remain healthy because all three of the above enzymes can metabolize the substrate, as they are from the same gene. C. The cell will remain healthy because the enzyme C mRNA will undergo alternative splicing again until it transformed into enzyme A mRNA. D. The cell will remain healthy because enzyme-substrate interactions are nonspecific and enzyme C will eventually metabolize the substrate.

A. The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown.

Different polysaccharides are used by plants for energy storage and structural support. The molecular structures for two common polysaccharides are shown in Figure 1. Starch is used by plants for energy storage, and cellulose provides structural support for cell walls. The monomer used to construct both molecules is glucose. A study determined the effect of two different digestive enzymes, A and B, on these two polysaccharides. Table 1 presents the data from the study. Mammals do not produce digestive enzyme B. However, sheep and cattle are two types of mammals that contain microorganisms in their digestive tract that produce enzyme B. Based the information provided, which of the following statements best describes why starch and cellulose provide different functions in plants? A. The differences in the assembly and organization of the monomers of these two polymers result in different chemical properties. B. Since starch and cellulose are composed of identical monomers, the cellular environment where they are located controls their function. C. The monomers of cellulose are connected by covalent bonds, making it idea for structural support. D. The monomers of starch are connected by ionic bonds, making it ideal for energy storage for plants.

A. The differences in the assembly and organization of the monomers of these two polymers result in different chemical properties.

Fibroblast growth factor receptors (FGFRs) are transmembrane proteins that regulate cellular processes such as cell proliferation and differentiation. The extracellular domains of FGFR proteins bind specifically to signaling molecules called fibroblast growth factors. The intracellular domains of FGFR proteins function as protein kinases, enzymes that transfer phosphate groups from ATP to protein substrates. FGFR activation occurs when binding by fibroblast growth factors causes FGFR proteins in the plasma membrane to become closely associated with each other. The association of two FGFR proteins stimulates protein kinase activity, which triggers the activation of intracellular signaling pathways. A simplified model of FGFR activation is represented in Figure 1. Figure 1. A simplified model of FGFR activation Which of the following changes in the FGFR signaling pathway is most likely to result in uncontrolled cell proliferation? A. The irreversible association of FGFR proteins B. The loss of the FGFR protein kinase function C. A decrease in the intracellular concentration of ATP D. A decrease in the extracellular concentrations of fibroblast growth factors

A. The irreversible association of FGFR proteins

Adjacent plant cells have narrow channels called plasmodesmata that pass through the cell walls of the connected cells and allow a cytoplasmic connection between the cells. Which of the following statements best describes a primary function of plasmodesmata? A. They allow the movement of molecules from one cell to another, enabling communication between cells. B. They prevent the cell membrane from pulling away from the cell wall during periods of drought. C. They eliminate the need to produce signaling molecules and eliminate the need for cells to have receptors for signaling molecules. D. They increase the surface area available for attachment of ribosomes and thus increase protein synthesis.

A. They allow the movement of molecules from one cell to another, enabling communication between cells.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Mice have melanocytes in the skin on their ears and show a tanning response to UV radiation. Researchers were studying a mutant population of mice that do not show a tanning response. Genetic testing of these mutant mice showed that the pathway causing the production of α-MSH by keratinocytes in response to UV radiation was fully functional. Thus, the researchers claimed that the lack of tanning response was due a nonfunctional MC1R. Which of the following pieces of evidence would best support the researchers' claim above? A. When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased. B. When researchers viewed sections of mutant mouse ears under the microscope, they found melanocyte numbers comparable to nonmutant mice. C. When researchers exposed the mutant mice to UV radiation, the amount of POMC mRNA in keratinocytes did not change. D. When researchers exposed the mutant mice to UV radiation, the level of melanin production did not change.

A. When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased.

A gene is: A. a segment of DNA that codes for a protein B. a set of homologous chromosomes C. a molecule within DNA D. a type of pants

A. a segment of DNA that codes for a protein

Down's Syndrome occurs when an individual has: A. an extra chromosome 21 B. a missing chromosome C. an extra set of homologous D. haploid cells

A. an extra chromosome 21

The image illustrates what phase of mitosis A. anaphase B. teleophase C. metaphase D. prophase

A. anaphase

Which site of the tRNA molecule binds to the mRNA molecule? A. anticodon B. codon C. amino acid D. 5 prime end

A. anticodon

The purpose of mitosis is to: A. create new cells B. copy DNA C. cellular repair D. reproduction

A. create new cells

The two sides of the DNA ladder are loosely bonded together by: A. hydrogen bonds B. molecular glue C. magnetism D. polymers

A. hydrogen bonds

DNA is called the "blueprint of life" because: A. it contains the plans for building an organism B. it is like a fingerprint C. it can relay messages to other cells D. it has a bright blue color

A. it contains the plans for building an organism

Okazaki fragments form on the: A.lagging strand B.leading strand C.base-pairs D. 5' end

A. lagging strand

Which of the following takes the genetic code from the nucleus to the cytoplasm? A. mRNA B. tRNA C. DNA D. ribonucleotides

A. mRNA

Before the Hershey-Chase experiment, many scientists believed that ____ carried the hereditary information. A. proteins B. phages C. helicase D. the plasma membrane

A. proteins

In the Hershey Chase Experiment, DNA was labeled with ____, and bacteriophage protein was labeled with _____. A. radioactive phosphorous, radioactive sulfar B. radioactive sulfur, radioactive phosphorous C. codons, anticodons D. DNA Polymerase, RNA polymerase

A. radioactive phosphorous, radioactive sulfur

Which of the following cells undergo meiosis? A. sperm cell B. sliver cells C. unicellular organisms D. all of these

A. sperm cells

Cytokinesis begins during which phase? A. telophase B. synthesis phase C. anaphase D. metaphase

A. telophase

Adenine always pairs with: A. thymine B. cytosine C. guanine D. ribose

A. thymine

Amino acids are joined together into a protein chain by which of the following? A. transfer RNA B. DNA polymerase C. hydrogen bonds D. messenger RNA

A. transfer RNA

RNA is synthesized on a DNA template in a process called ______, which utilizes the enzyme _______ A. translation, RNA polymerase B. transcription, DNA polymeraset C. ranscription, RNA polymerase D. replication, DNA polymerase

A. translation, RNA polymerase

A cell that has 20 chromosomes undergoes mitosis. Which of the following is true? A. two daughter cells will be created, each have 20 chromosomes B. two daughter cells will be created, each have 40 chromosomes C. 4 daughter cells will be created, each having 10 chromosomes D. 2 daughter cells will be created, each having 10 chromosomes

A. two daughter cells will be created, each have 20 chromosomes

Which of the following is found on RNA but not DNA? A. uracil B. deoxyribose C. phosphate D. adenine

A. uracil

The sugar found in DNA is: A. equal B. deoxyribose C. ribose D. glucose

B. . deoxyribose

Students carry out a genetics experiment to investigate the inheritance pattern of the white-eye trait in fruit flies. In the experiment, the students cross a red-eyed female with a white-eyed male to produce an F1 generation. The students observe that all the flies in the F1 generation have red eyes. The students then allow the F1 flies to mate and produce an F2 generation. The students will use the F2 data to perform a chi-square goodness-of-fit test based on a null hypothesis of autosomal recessive inheritance. The observed and expected values for the chi-square goodness-of-fit test are shown in Table 1. Table 1. The observed and expected values for a chi-square goodness-of-fit test The students plan to use a significance level of p=0.01. Which of the following is the most appropriate critical value for the students to use in their chi-square goodness-of-fit test? A. 7.82 B. 11.34 C. 13.28 D. 326.7

B. 11.34

A cell with a diploid number of 24 undergoes meiosis, how many chromosomes are in each daughter cell? A. 6 B. 12 C. 24 D. 48

B. 12

DNA replication results in: A. 2 completely new DNA molecules B. 2 DNA molecules that each contain a strand of the original C. 1 new DNA molecule, 1 old molecule is conserved D. 1 new molecule of RNA

B. 2 DNA molecules that each contain a strand of the original

How many different amino acids are used to assemble proteins in cells? A. 3 B. 20 C. 100 D. an infinite number

B. 20

Four trials measuring recombination frequency between gene R and gene L were conducted, and the results are shown in Table 1. Table 1. Recombination frequency The mean map distance between gene R and gene L is closest to which of the following? A. 0.28 map units B. 28 map units C. 0.14 map units D. 14 map units

B. 28 map units

Two fruit fly mutations are ebony body (e) and sepia eyes (s). Four different students performed dihybrid crosses with flies that were heterozygous with a mutant allele and a wild-type allele for both genes (EeSs×EeSs). The results are shown in Table 1. Table 1. Offspring of four separate dihybrid crosses The mean number of fruit flies per student that are homozygous recessive for both genes is closest to which of the following? A. 89.75 B. 29.0 C. 22.75 D. 18.5

B. 29.0

B. 3

Meiosis results in _____ A. 2 haploid daughter cells B. 4 haploid daughter cells C. 2 diploid daughter cells D. 4 diploid daughter cells

B. 4 haploid daughter cells

Sex chromosomes determine the phenotype of sex in humans. Embryos containing XX chromosomes develop into females, and embryos containing XY chromosomes develop into males. The sex chromosomes separate during meiosis, going to different gamete cells. A woman is heterozygous for the X-linked recessive trait of hemophilia A. Her sex chromosomes can be designated as XHXh. During meiosis the chromosomes separate as shown in Figure 1. Figure 1. Transmission pattern for sex chromosomes of a woman heterozygous for hemophilia A into gametes If the woman and a man with normal clotting function have children, what is the probability of their children exhibiting hemophilia A? A. 50 percent for daughters, 0 percent for sons B. 50 percent for sons, 0 percent for daughters C. 50 percent for all children D. 0 percent for all children

B. 50 percent for sons, 0 percent for daughters

Researchers performed a dihybrid cross with coffee bean plants to investigate whether the inheritance of two traits (height and stem circumference) follows Mendel's law of independent assortment. The data for the F2 generation are presented in Table 1. Table 1. Data for the F2 generation Which of the following is closest to the calculated chi-square (χ2) value for the data presented in Table 1? A. 8.35 B. 72.01 C. 98.00 D. 2,546.00

B. 72.01

The base pair rules states that: A. Replication is semiconservative B. A pairs with T, G pairs with C C. DNA is a double helix held together by hydrogen bonds D. A pairs with G, T pairs with C

B. A pairs with T, G pairs with C

Given the following DNA strand, which of the following is its complementary mRNA? G G A C T G A T T A. C C T G A C T A A B. C C U G A C U A A C. G G A C T G A T T D. T T A G T C A G G

B. CCUGACUAA

A student wants to modify model 1 so that it represents an RNA double helix instead of a DNAdouble helix. Of the following possible changes, which would be most effective in making model 1 look more like RNA than DNA? A. Changing the sequence of the base pairs B. Changing the deoxyriboses to riboses by adding −OH groups C. Changing the shapes of the nitrogenous bases to match those shown in model 2 D. Changing the sugar-phosphate backbone to a ribbon, as shown in model 3

B. Changing the deoxyriboses to riboses by adding −OH groups

Which of the following must happen first in order for DNA replication to occur? A.DNA polymerase binds to the leading strand B. DNA is unwound C.Hydrogen bonds form between bases D. chromosomes condense

B. DNA is unwound

The Hershey-Chase research showed that: A. bacteria could be transformed B. DNA was the molecule of heredity C.DNA was in the shape of a double helix D. bases paired in a predictable way (A's to T's and G's to C's)

B. DNA was the molecule of heredity

Signal transduction may result in changes in gene expression and cell function, which may alter phenotype in an embryo. An example is the expression of the SRY gene, which triggers the male sexual development pathway in mammals. This gene is found on the Y chromosome. Which statement provides the evidence to justify the claim that signal transduction may result in an altered phenotype? A. If the SRY gene is absent or nonfunctional, the embryo will exhibit male sexual development. B. If the SRY gene is absent or nonfunctional, the embryo will exhibit female sexual development. C. An embryo with a male sex chromosome will always exhibit male sexual development. D. An embryo with two male sex chromosomes will always exhibit male sexual development.

B. If the SRY gene is absent or nonfunctional, the embryo will exhibit female sexual development.

A person's blood glucose level fluctuates during the day, as represented in Figure 1. Two hormones, insulin and glucagon, are directly involved in regulating the blood glucose level to maintain a healthy level. Insulin acts to lower the blood glucose level, and glucagon acts to increase the blood glucose level.Figure 1. Blood glucose fluctuations of an individual. Which of the following best predicts what will happen to the blood glucose level if the person has another meal at 5 p.m.? A. Immediately after the meal, the blood glucose level will decrease because of the increase in glucagon levels. B. Immediately after the meal, the blood glucose level will increase, and then insulin will be secreted to counter the increase. C. Several hours after the meal, the blood glucose level will increase sharply because of an increase in the amount of glucagon secreted. D. The blood glucose level will not change after the 5 p.m. meal because the person has already consumed two meals and the blood glucose level has been adjusted to a steady-state level.

B. Immediately after the meal, the blood glucose level will increase, and then insulin will be secreted to counter the increase.

The epidermal growth factor receptor EGFR is a cell surface receptor. When a growth factor binds to EGFR, the receptor is activated. The activated EGFR triggers a signal transduction pathway, which leads to increased frequency of cell division. Which of the following best predicts the effect of a mutation that causes EGFR to be active in the absence of a growth factor? A. Increased apoptosis will lead to abnormal growth of the tissue. B. Increased cell division will lead to the formation of a tumor. C. Cells will exit the cell cycle, entering a non-dividing G0 phase. D. Fewer cells will be in any of the stages of mitosis.

B. Increased cell division will lead to the formation of a tumor.

Metformin is a drug used to treat type 2 diabetes by decreasing glucose production in the liver. AMP-activated protein kinase (AMPK) is a major cellular regulator of glucose metabolism. Metformin activates AMPK in liver cells but cannot cross the plasma membrane. By blocking AMPK with an inhibitor, researchers found that AMPK activation is required for metformin to produce an inhibitory effect on glucose production by liver cells. Which of the following best describes the component that metformin represents in a signal transduction pathway that regulates glucose production in the liver? A. It is a secondary messenger that amplifies a signal through a cascade reaction. B. It is a ligand that activates the signal transduction pathway of the activation of AMPK. C. It is an allosteric regulator that binds to a crucial section of the DNADNA that makes the enzymes needed for glucose uptake. D. It is a competitive inhibitor that binds to glucose and prevents it from entering the cell.

B. It is a ligand that activates the signal transduction pathway of the activation of AMPK.

When a mustard plant seedling is transferred to an environment with higher levels of carbon dioxide, the new leaves have a lower stomata-to-surface-area ratio than do the seedling's original leaves. Which of the following best explains how the leaves from the same plant can have different stomatal densities when exposed to an elevated carbon dioxide level? A. Increased photosynthesis leads to larger leaves that need more stomata for photosynthesis, leading to an increase in stomatal density. B. Leaf growth is promoted through increased photosynthesis, but the genetically regulated rate of stomatal production is not altered, leading to a decrease in stomatal density. C. Leaf growth is inhibited by decreased photosynthesis, and the genetically regulated rate of stomatal production remains the same, leading to an increase in stomatal density. D. Leaf growth is inhibited by decreased photosynthesis, and the genetically regulated rate of stomatal production remains the same, leading to a decrease in stomatal density.

B. Leaf growth is promoted through increased photosynthesis, but the genetically regulated rate of stomatal production is not altered, leading to a decrease in stomatal density.

Phosphofructokinase (PFK) is a key enzyme in glycolysis. ATP is one of the two substrates for the reaction catalyzed by PFK. ATP is also an allosteric regulator of PFK. Figure 1 shows the enzyme-substrate interactions of PFK. Figure 1. The enzyme-substrate interactions of PFK A researcher found a mutation that resulted in the PFK enzyme being unable to bind ATP to the allosteric site. Which of the following best predicts the effect of the mutation? A. The activity of the enzyme will not be affected because the active site is not involved in substrate binding at the allosteric site. B. Negative feedback regulation does not occur, so the enzyme will be active when glycolysis is not needed. C. Positive feedback does not occur, and the activity of the enzyme will decrease when glycolysis is needed. D. The activity of the enzyme will fluctuate independent of the ATP concentration.

B. Negative feedback regulation does not occur, so the enzyme will be active when glycolysis is not needed.

In pea plants, purple flower color is dominant to red flower color and long pollen grains are dominant to round pollen grains. Researchers crossed two pure-breeding lines of the pea plants to investigate whether the genes controlling flower color and pollen shape segregate independently. The procedure for the genetics experiment is summarized in Figure 1. Figure 1. Summary of a genetics experiment using pea plants Which of the following tables best shows the expected values in the F2 generation for a chi-square goodness-of-fit test for a model of independent assortment? A. PhenotypeExpectedPurple, long533Purple, round533Red, long533Red, round533 B. PhenotypeExpectedPurple, long1199Purple, round400Red, long400Red, round133 C. PhenotypeExpectedPurple, long1067Purple, round355Red, long355Red, round355 D. PhenotypeExpectedPurple, long1002Purple, round515Red, long515Red, round1002

B. PhenotypeExpectedPurple, long1199Purple, round400Red, long400Red, round133

Glycolysis is a metabolic pathway that converts glucose into pyruvate and is observed in each of the three domains. The hexokinase family of enzymes is required during glycolysis to phosphorylate six-carbon sugars. Researchers designed a general hexokinase inhibitor that is effective in the neurons of rats. Which of the following best predicts the effect of adding this inhibitor to a culture of plant cells? A. Plant cells will be unaffected by the inhibitor as they do not perform glycolysis. B. Plant cells will be unable to perform glycolysis due to the inhibitor and will die. C. Plant cells will be unable to perform photosynthesis due to the inhibitor and will die. D. Plant cells will still be able to perform glycolysis since plant hexokinase is not structurally similar to animal hexokinase.

B. Plant cells will be unable to perform glycolysis due to the inhibitor and will die.

Huntington's disease, an autosomal dominant disorder, is caused by a mutation in the HTT gene. The HTT gene contains multiple repeats of the nucleotide sequence CAG. A person with fewer than 35 CAG repeats in the HTT gene is unlikely to show the neurological symptoms of Huntington's disease. A person with 40 or more CAG repeats almost always becomes symptomatic. Due to errors in meiosis, an individual without symptoms of Huntington's disease can produce gametes with a larger number of CAG repeats than there are in their somatic cells. A woman develops Huntington's disease. Her father had the disorder. Her mother did not, and there is no history of the disorder in the mother's family. Which of the following best explains how the woman inherited Huntington's disease? A. She inherited an allele with fewer than 40 CAG repeats in the HTT gene because her mother did not have Huntington's disease. B. She inherited an allele with more than 40 CAG repeats in the HTT gene from her father. C. Her mother produced eggs that all have more than 40 repeats in the HTT gene. D. Her mother produced eggs that all have fewer than 40 CAGCAG repeats in the HTT gene.

B. She inherited an allele with more than 40 CAG repeats in the HTT gene from her father.

Figure 1 represents a portion of a process that occurs during protein synthesis. Figure 1. Model of selected features of DNA transcription Which claim is most consistent with the information provided by the diagram and current scientific understanding of gene regulation and expression? A. Reversible changes in the DNA sequence may influence how a gene is expressed in a cell. B. Some sequences of DNA can interact with regulatory proteins that control transcription. C. This is an inducible operon controlled by several regulatory factors. D. The transcription factor may produce mutations in the binding site at the promoter sequence inhibiting the synthesis of the protein.

B. Some sequences of DNA can interact with regulatory proteins that control transcription.

Different polysaccharides are used by plants for energy storage and structural support. The molecular structures for two common polysaccharides are shown in Figure 1. Starch is used by plants for energy storage, and cellulose provides structural support for cell walls. The monomer used to construct both molecules is glucose. A study determined the effect of two different digestive enzymes, A and B, on these two polysaccharides. Table 1 presents the data from the study. Mammals do not produce digestive enzyme B. However, sheep and cattle are two types of mammals that contain microorganisms in their digestive tract that produce enzyme B. Based on Figure 1, which of the following best compares the atomic structures of starch and cellulose? A. Starch is composed of carbon, hydrogen, and oxygen, while cellulose also contains nitrogen. B. Starch and cellulose are composed of repeating glucose monomers; however, in cellulose every other glucose monomer is rotated 180 degrees. C. Starch is composed of monomers that each have a CH2OHCH group, while cellulose only has a CH2OHCH group on every other monomer. D. Starch and cellulose are composed of identical monomers and therefore have identical structures.

B. Starch and cellulose are composed of repeating glucose monomers; however, in cellulose every other glucose monomer is rotated 180 degrees.

The relative amounts of DNA present in the nucleus of a cell at four different stages of the life cycle are shown in Figure 1. Figure 1. Relative amounts of DNA present in the nucleus of a cell Based on Figure 1, which of the following statements correctly links a stage of the cell cycle with the event occurring at that stage? A. Stage II represents the G2 phase of the cell cycle. B. Synthesis of sufficient DNA for two daughter cells occurs in stage II. C. Stage III includes mitosis. D. The replication of genetic material occurs in stage IV.

B. Synthesis of sufficient DNA for two daughter cells occurs in stage II.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Which of the following claims about the TYR, TRP2, and TRP1 mammalian genes is most likely to be accurate? A. The TYR, TRP2, and TRP1 genes are located next to each other on a single chromosome and are organized into an operon. B. The TYR, TRP2, and TRP1 genes may be located on different chromosomes but are activated by the same transcription factor. C. The TYR, TRP2, and TRP1 genes are identical genes since they are activated by the same transcription factor. D. The TYR, TRP2, and TRP1 genes may be located on different chromosomes but with identical operator sequences.

B. The TYR, TRP2, and TRP1 genes may be located on different chromosomes but are activated by the same transcription factor.

Lynch syndrome is an inherited condition associated with an increased risk for colon cancer, as well as certain other cancers. Mutations in one of several genes involved in DNA repair during DNA replication have been associated with Lynch syndrome. DNA sequencing was performed for an individual. The results indicated that the individual carries one of the dominant alleles that has been associated with Lynch syndrome. Which of the following best explains how the results should be interpreted? A. The individual does not have an increased risk of developing cancer because one dominant allele is insufficient to cause the disease. B. The individual has an increased risk of developing colon cancer. C. Because the person's DNA has the mutation, other family members must have cancer. D. Results cannot be interpreted until testing determines if additional mutated alleles are present.

B. The individual has an increased risk of developing colon cancer.

A true-breeding variety of wheat that produces deep-red-colored grain was crossed with a true-breeding variety that produces a white-colored grain. The resulting F1 offspring all had medium-red-colored grain. Five crosses of the F1 plants were all grown under the same conditions and resulted in plants with a variety of grain color, as indicated in Table 1. Table 1. F2 phenotypes resulting from five crosses Which of the follow indicates the mean number per cross of F2 plants producing medium-red grain and correctly explains the distribution of the phenotypes? A. The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that grain color is under environmental control. B. The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that multiple genes are involved in grain color determination. C. The mean number of medium-red phenotypes per cross is 104. The distribution of phenotypes suggests that grain color is under environmental control. D. The mean number of medium-red phenotypes per cross is 104. The distribution of phenotypes suggests that multiple genes are involved in grain color determination.

B. The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that multiple genes are involved in grain color determination.

The insulin receptor is a transmembrane protein that plays a role in the regulation of glucose homeostasis. The receptor's extracellular domain binds specifically to the peptide hormone insulin. The receptor's intracellular domain interacts with cellular factors. The binding of insulin to the receptor stimulates a signal transduction pathway that results in the subcellular translocation of GLUT4, a glucose transport protein that is stored in vesicles inside the cell. A simplified model of the insulin receptor-signaling pathway is shown in Figure 1. Figure 1. A simplified model of the insulin receptor-signaling pathway Which of the following statements best predicts the effect of a loss of function of the insulin receptor's intracellular domain? A. The stimulation of the signal transduction pathway will increase. B. The storage of GLUT4 in vesicles inside the cell will increase. C. The number of GLUT4 molecules in the plasma membrane will increase. D. The concentration of glucose inside the cell will increase.

B. The storage of GLUT4 in vesicles inside the cell will increase.

Figure 1 represents the relative time and sequence of the phases of the cell cycle. Figure 1. Representation of the cell cycle and identification of the G1/S checkpoint Which statement best predicts why a cell's progression through the cell cycle might be halted at the G1/S checkpoint? A. Spindle fibers have not correctly attached to chromosomes. B. There are not enough nucleotides available to construct new DNA. C. Damage occurred to DNA when it was being copied in G1. D. Proteins necessary for MM phase of the cell cycle have not been produced.

B. There are not enough nucleotides available to construct new DNA.

The three nucleotides found in DNA will code for a single: A. protein B. amino acid C. messenger RNA D. sugar

B. amino acid

Which of the following best describes the arrangement of the sides of the DNA molecule? A. twisted B. antiparallel C. bonded D. alternating

B. antiparallel

What did Frederick Griffith's discover? A. DNA carried genetic information B. bacteria could transform C. mice died when exposed to viruses D. viruses were encapsulated

B. bacteria could transform

In order to transform to a virulent form of bacteria, non encapsulated bacteria must: A. divide B. be exposed to killed capsulated bacteria C. be exposed to radioactive phosphorous D. infect a host

B. be exposed to killed capsulated bacteria

The 5' and 3' numbers are related to the: A. length of the DNA strand B. carbon rings in sugar C. the number of phosphates D. the base pair rule

B. carbon rings in sugar

The three nucleotide sequence on RNA is called a: A. tRNA B. codon C. triplet D. gene

B. codon

The process of mitosis ensures that: A. each new cell is genetically different from its parent B. each new cell receives the proper number of chromosomes C. cells will divide at the appropriate time D. DNA is replicated without errors

B. each new cell receives the proper number of chromosomes

Transcription results in: A. an amino acid chain B. messenger RNA C. complementary DNA D. okazaki fragments

B. messenger RNA

Compared to the X chromosome, the Y chromosome is: A. much larger B. much smaller C. more twisted D. inherited more often

B. much smaller

Okazaki fragments occur during: A. transformation B. replication C. polymerase reaction D. synthesis

B. replication

The process by which DNA makes a copy of itself is called: A. synthesis B. replication C. transcription D. translation

B. replication

The sugar in RNA is _____, the sugar in DNA is _______ A. deoxyribose, ribose B. ribose, deoxyribose C. ribose, phosphate D. ribose, uracil

B. ribose, deoxyribose

The process by which RNA is made from DNA is called: A. translation B. transcription C. replication D. pagination

B. transcription

The two men who established the structure of DNA were: A. Frederick and Alvers B. Watson and Crick C. Berkely and Fry D. Darwin and Lamarke

B. watson and crick

A student used microscopy to investigate the relative lengths of the different stages of mitosis. The student prepared slides of cells isolated from a growing onion root tip and viewed the slides under a dissecting microscope. The student then made diagrams of cells that were in different stages of mitosis and counted the number of cells that were in each of those stages. The student's data are presented in Table 1. Table 1. Number of cells in each of four different stages of mitosis Based on the data, the percent of the mitotic cells that were in metaphase is closest to which of the following? A. 5% B. 11% C. 18% D. 66%

C. 18%

A model of the typical life cycle of a cell is shown in Figure 1. Figure 1. Typical life cycle of a eukaryotic cell Scientists have estimated that it takes yeast cells approximately 20 hours to complete the entire cycle. Table 1 shows the amount of time in each phase of the life cycle for yeast cells. Based on Table 1, what percent of the life cycle of yeast cells is spent in DNA replication? A. 5 percent B. 10 percent C. 25 percent D.50 percent

C. 25 percent

How many different codons are possible? A. 3 B. 20 C. 64 D. an infinite number

C. 64

Which of the following steps in a signaling pathway typically occurs first once a chemical messenger reaches a target cell? A. Specific genes are activated. B. A second messenger molecule is produced. C. A ligand binds to a receptor. D. Specific proteins are synthesized.

C. A ligand binds to a receptor.

Different polysaccharides are used by plants for energy storage and structural support. The molecular structures for two common polysaccharides are shown in Figure 1. Starch is used by plants for energy storage, and cellulose provides structural support for cell walls. The monomer used to construct both molecules is glucose. A study determined the effect of two different digestive enzymes, A and B, on these two polysaccharides. Table 1 presents the data from the study. Mammals do not produce digestive enzyme B. However, sheep and cattle are two types of mammals that contain microorganisms in their digestive tract that produce enzyme B. Which of the following best describes the process that adds a monosaccharide to an existing polysaccharide? A. The monosaccharide is completely broken down by a specific enzyme and then the atoms are reorganized and made into a polysaccharide. B. Ionic bonds are formed between adjacent carbon atoms of the monosaccharide and the polysaccharide by adding water (H2O) and a specific enzyme. C. A specific enzyme removes the hydrogen (H) from the monosaccharide and the hydroxide (OH) from the polysaccharide, creating a bond between the two and creating a water (H2O) molecule. D. A specific enzyme removes two hydroxides (OH), one from the monosaccharide, and one from the polysaccharide, creating a bond between the two monosaccharides and creating a hydrogen peroxide (H2O2) molecule.

C. A specific enzyme removes the hydrogen (H) from the monosaccharide and the hydroxide (OH) from the polysaccharide, creating a bond between the two and creating a water (H2O) molecule.

Huntington's disease has been traced to the number of CAG repeats in the HTT gene, which is located on chromosome 4. The phenotypic influence of individual alleles with different numbers of repeats is shown in Table 1. Table 1: Effect of CAG repeat number on Huntington's disease expression CAG RepeatsExpression<27No symptoms27-35No symptoms but increased risk for offspring36-39May develop symptoms in old age>40Will express symptoms from an early age Which of the following is most likely the immediate cause of the first appearance of Huntington's disease in a person? A. A point mutation occurs in the HTT gene. B. The first appearance of the CAG repeat occurs in the HTT gene. C. An allele with more than 39 CAG repeats was inherited by the affected person. D. The person inherited two alleles that each contained 20 CAG repeats.

C. An allele with more than 39 CAG repeats was inherited by the affected person.

Figure 1. Four different bonds (W, X, Y, and Z) in a DNA molecule Figure 1 represents a segment of DNA. Radiation can damage the nucleotides in a DNA molecule. To repair some types of damage, a single nucleotide can be removed from a DNA molecule and replaced with an undamaged nucleotide. Which of the four labeled bonds in Figure 1 could be broken to remove and replace the cytosine nucleotide without affecting the biological information coded in the DNA molecule? A. Bond X only B. Bond W only C. Bonds Y and Z at the same time D. Bonds W and Z at the same time

C. Bonds Y and Z at the same time

Figure 1. Molecule 1 represents RNA, and molecule 2 represents DNA. Which of the following best describes a structural similarity between the two molecules shown in Figure 1 that is relevant to their function? A. Both molecules are composed of the same four nucleotides, which allows each molecule to be produced from the same pool of available nucleotides. B. Both molecules are composed of the same type of five-carbon sugar, which allows each molecule to act as a building block for the production of polysaccharides. C. Both molecules contain nucleotides that form base pairs with other nucleotides, which allows each molecule to act as a template in the synthesis of other nucleic acid molecules. D. Both molecules contain nitrogenous bases and phosphate groups, which allows each molecule to be used as a monomer in the synthesis of proteins and lipids.

C. Both molecules contain nucleotides that form base pairs with other nucleotides, which allows each molecule to act as a template in the synthesis of other nucleic acid molecules.

The element that transformed the bacteria in Griffith's experiments was: A.protein B. capsules C. DNA D. heat

C. DNA

Cystic fibrosis (CF) is a progressive genetic disease that causes persistent lung infections and affects the ability to breathe. CF is inherited in an autosomal recessive manner, caused by the presence of mutations in both copies of the gene for the cystic fibrosis transmembrane conductance regulator (CFTR) protein. Partial nucleotide sequences and the corresponding amino acid sequences for an unaffected individual and an affected individual are modeled in Figure 1. Figure 1. CFTR protein sequences in unaffected and affected individuals Based on the information in Figure 1, which type of mutation explains the nature of the change in DNA that resulted in cystic fibrosis in the affected individual? A. Substitution, because the amino acid tryptophan is replaced with glycine. B. Insertion, because an extra guanine is present, which changes the reading frame. C. Deletion, because a thymine is missing, which changes the reading frame. D. Duplication, because the amino acid leucine occurs twice, which changes the reading frame

C. Deletion, because a thymine is missing, which changes the reading frame.

Scientists conducted a transformation experiment using E. coli bacteria and the pTru plasmid. Samples of the pTru plasmid (lane A) and the chromosomal DNA from two different E. coli strains that the scientists attempted to transform (lane B and lane C) were compared using gel electrophoresis. The results are shown in Figure 1. Figure 1. Results of E. coli transformation with pTru plasmid Which of the following statements best explains the experimental results observed in Figure 1 ? A. E. coli in both lanes BB and CC have been successfully transformed and contain additional genetic information. B. E. coli in lane BB have been successfully transformed and contain additional genetic information. C. E. coli in lane CC have been successfully transformed and contain additional genetic information. D. Which E. coli have been transformed cannot be determined from this gel.

C. E. coli in lane CC have been successfully transformed and contain additional genetic information.

Nucleotide base pairing in DNA is universal across organisms. Each pair (T−A; C−G) consists of a purine and a pyrimidine. Which of the following best explains how the base pairs form? A. Ionic bonds join two double-ringed structures in each pair. B. Hydrogen bonds join two single-ringed structures in each pair. C. Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair. D. Covalent bonds join a double-ringed structure to a single-ringed structure in each pair.

C. Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair.

Figure 1 depicts a simplified model of a replication bubble. Figure 1. DNA replication bubble. Which of the following best explains how this model illustrates DNA replication of both strands as a replication fork moves? A. II and IV are synthesized continuously in the 5′5′ to 3′3′ direction. B. II and III are synthesized in segments in the 3'3′ to 5'5′ direction. C. I is synthesized continuously in the 5' to 3′ direction, and III is synthesized in segments in the 5′ to 3′ direction. D. II is synthesized in segments after DNA polymerase is released from synthesizing strand IV.

C. I is synthesized continuously in the 5′ to 3′direction, and III is synthesized in segments in the 5′ to 3′ direction.

The tadpoles of Mexican spadefoot toads are known to exhibit phenotypic plasticity depending on food availability. Tadpole mouthparts can vary significantly, prompting researchers to categorize them as either omnivore-morph or carnivore-morph. Carnivore-morph tadpoles are larger and have mouthparts that are better suited for predation. Remarkably, carnivore-morph tadpoles can change into omnivore-morph tadpoles when the food supply changes. Which of the following best describes an advantage of the phenotypic plasticity displayed by the tadpoles? A. It allows the tadpoles to change their genome in response to environmental pressures. B. It enables the tadpoles to develop into a distinct species of toads. C. It gives the tadpoles increased versatility with respect to diet. D. It allows the tadpoles to delay metamorphosis until there is maximal food available for the adults.

C. It gives the tadpoles increased versatility with respect to diet.

C. Keratinocytes do not express the MITF gene.

R. C. Punnett conducted experiments on the inheritance of traits in the sweet pea, Lathyrus odoratus. In one experiment, he crossed two different true-breeding sweet pea plant strains, one with erect petals and long pollen, and the other with hooded petals and round pollen. All the offspring (F1 generation) had erect petals and long pollen (Figure 1). Figure 1. Sweet pea plant cross Next, Punnett allowed the F1 generation to self-fertilize and recorded the phenotypes of their offspring. The data are shown in Table 1. Sweet pea plants have a diploid (2n) chromosome number of 14. Which of the following best explains how the sweet pea plants in the parental generation produce F1 offspring with 14 chromosomes? A. Meiosis I and II lead to the formation of cells with 14 chromosomes. When two cells combine during fertilization, extra chromosomes are randomly broken down, leading to offspring with 14 chromosomes. B. Meiosis I and II lead to the formation of cells with 14 chromosomes. When two cells combine during fertilization, extra chromosomes with recessive traits are broken down, leading to offspring with 14 chromosomes. C. Meiosis I and II lead to the formation of cells with 7 chromosomes. During meiosis II, homologous chromosomes separate. During meiosis II, sister chromatids separate. Two cells combine during fertilization to produce offspring with 14 chromosomes. D. Meiosis I and II lead to the formation of cells with 7 chromosomes. During meiosis II, sister chromatids separate. During meiosis II, homologous chromosomes separate. Two cells combine during fertilization to produce offspring with 14 chromosomes.

C. Meiosis I and II lead to the formation of cells with 7 chromosomes. During meiosis II, homologous chromosomes separate. During meiosis II, sister chromatids separate. Two cells combine during fertilization to produce offspring with 14 chromosomes.

Which of the following is common feature of the illustrated reactions showing the linking of monomers to form macromolecules? A. Two identical monomers are joined by a covalent bond. B. Two different monomers are joined by a covalent bond. C. Monomers are joined by a covalent bond, and a water molecule is produced. D. Monomers are joined by ionic bonds, and a water molecule is produced.

C. Monomers are joined by a covalent bond, and a water molecule is produced.

All cells must transcribe rRNA in order to construct a functioning ribosome. Scientists have isolated and identified rRNA genes that contribute to ribosomal structure for both prokaryotes and eukaryotes. Figure 1 compares the transcription and processing of prokaryotic and eukaryotic rRNA. Figure 1. Comparison of rRNA processing in prokaryotes and eukaryotes Which of the following statements provides the best explanation of the processes illustrated in Figure 1 ? A. Introns are removed from the pre-rRNA, and the mature rRNA molecules are joined and then translated to produce the protein portion of the ribosome. B. Introns are removed from the pre-rRNA, and each mature rRNA molecule is translated to produce the proteins that make up the ribosomal subunits. C. Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits. D. Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to bring different amino acids to the ribosome.

C. Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits.

For sexually reproducing diploid parent cells, which of the following statements best explains the production of haploid cells that occurs in meiosis but not in mitosis? A. Separation of chromatids occurs once, and there is one round of cell division in meiosis. B. Separation of chromatids occurs twice, and there are two rounds of cell division in mitosis. C. Separation of chromatids occurs once, and there are two rounds of cell division in meiosis. D. Separation of chromatids occurs twice, and there is one round of cell division in mitosis.

C. Separation of chromatids occurs once, and there are two rounds of cell division in meiosis.

Which of the following describes a key difference among the 20 amino acids that are used to make proteins? A. Only some amino acids have an R-group. B. Only some amino acids have a carboxyl group (COOH). C. Some amino acids are hydrophobic. D. Some amino acids contain the element phosphorus.

C. Some amino acids are hydrophobic.

An evolutionary biologist hypothesizes that two morphologically similar plant species are not closely related. To test the hypothesis, the biologist collects DNA samples from each of the two plant species and then uses restriction enzymes to cut the DNA samples into fragments, which are then subjected to gel electrophoresis. The results are shown in Figure 1. Figure 1. DNA analysis of two species Given the results shown in Figure 1, which of the following correctly describes a relationship between the two species? A. Species B is the ancestor of species A because it has fewer bands. B. Species A is more complex than species B because it has more bands. C. Species B has more short fragments of DNA than species A does. D. Species A has more short fragments of DNA than species B does.

C. Species B has more short fragments of DNA than species A does.

A DNA strand has the following bases: A A G C C A. What are the bases on its complimentary strand? A. A A G C C A B. A C C G A A C. T T C G G T D. C C A T T C

C. TTCGGT

A hydrophilic peptide hormone is produced in the anterior pituitary gland located at the base of the brain. The hormone targets specific cells in many parts of the body. Which of the following best explains a possible mechanism that would enable the hormone to efficiently reach all of the target cells in the body? A. The hormone interacts with the nerves at the base of the brain and directs signals to the target cells through the nervous system. B. The hormone diffuses into target cells adjacent to the anterior pituitary gland, where the hormone is degraded. C. The hormone is released into the bloodstream where it can be transported to all cells with the correct receptors. D. The hormone moves through cytoplasmic connections between cells until it has reached all cells with the correct intracellular binding sites.

C. The hormone is released into the bloodstream where it can be transported to all cells with the correct receptors.

The figure shows the results of an experiment to investigate the effects of an enriched CO2 environment on plant growth. Identical plants were separated into different groups and grown either in a standard CO2 environment (400 ppm CO2) or in an enriched CO2 environment (700 ppm CO2). Of the plants in each environment, half were grown under ideal conditions and half were grown under stressed conditions. Based on the figure, which statement best describes the observed relationship between atmospheric CO2 enrichment and plant growth under ideal and stressed conditions? A. The increase in atmospheric CO2 had no observable effect on plant growth under either ideal or stressed conditions. B. The increase in atmospheric CO2 resulted in a greater increase in plant growth under ideal conditions than under stressed conditions. C. The increase in atmospheric CO2 resulted in a greater increase in plant growth under stressed conditions than under ideal conditions. D. The increase in atmospheric CO2 resulted in an inhibition of plant growth under both ideal and stressed conditions.

C. The increase in atmospheric CO2 resulted in a greater increase in plant growth under stressed conditions than under ideal conditions.

Phosphorous (P) is an important nutrient for plant growth. Figure 1 shows Arabidopsis thaliana plants grown under phosphorus‐sufficient (left) and phosphorus‐starved (right) conditions for six weeks. Figure 1. Arabidopsis thaliana plants grown for six weeks. Which of the following is the most likely reason for the difference in leaf growth? A. The phosphorus-starved plant was unable to synthesize both the required proteins and lipids, limiting growth. B. The phosphorus-starved plant was unable to synthesize both the required proteins and carbohydrates, limiting growth. C. The phosphorus-starved plant was unable to synthesize both the required nucleic acids and lipids, limiting growth. D. The phosphorus-starved plant was unable to synthesize both the required carbohydrates and nucleic acids, limiting growth.

C. The phosphorus-starved plant was unable to synthesize both the required nucleic acids and lipids, limiting growth.

Retroviruses such as HIV and hepatitis B virus use RNA as their genetic material rather than DNA. In addition, they contain molecules of reverse transcriptase, an enzyme that uses an RNA template to synthesize complementary DNA. Which of the following best predicts what will happen when a normal cell is exposed to a retrovirus? A. The reverse transcriptase will cut the host DNA into fragments, destroying the host cell. B. The reverse transcriptase will insert the viral RNA into the host's genome so it can be transcribed and translated. C. The reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated. D. The reverse transcriptase will force the host ribosomes to translate the viral RNA prior to polypeptide assembly.

C. The reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated.

As shown in the diagram, when environmental temperatures drop below freezing, a layer of ice typically forms on the surface of bodies of freshwater such as lakes and rivers. Which of the following best describes how the structure of ice benefits the organisms that live in the water below? A. The water molecules in ice are closer together than those in liquid water, so the ice prevents the passage of air to the water, maintaining a constant gas mixture in the water. B. The water molecules in ice are closer together than those in liquid water, so the ice forms a barrier that protects the organisms in the water from the freezing air temperatures. C. The water molecules in ice are farther apart than those in liquid water, so the ice floats, maintaining the warmer, denser water at the lake bottom. D.The water molecules in ice are farther apart than those in liquid water, so the ice floats, preventing the escape of gases from the liquid water.

C. The water molecules in ice are farther apart than those in liquid water, so the ice floats, maintaining the warmer, denser water at the lake bottom.

Researchers tracked the amount of DNA (measured in picograms) over time beginning with a single cell and continuing through several rounds of cell division. The researchers observed threadlike chromosomes prior to cell division. The threadlike chromosomes disappeared from view shortly after each division. The amount of DNA in picograms per cell over several rounds of cell division is shown in Figure 1. Figure 1. Amount of DNA in picograms per cell over several rounds of cell division Which of the following statements is consistent with the data in Figure 1? A. The cells have a haploid chromosome number of 3. B. The cells have a diploid chromosome number of 6. C. There is a change from 3 to 6 picograms of DNA because DNA is replicated before each round of cell division. D. There is a change from 6 to 3 picograms of DNA after each cell division because the chromosomes lengthen following cell division. Answer D

C. There is a change from 3 to 6 picograms of DNA because DNA is replicated before each round of cell division.

The tumor suppressor protein p53 binds to DNA and activates target genes, which results in the synthesis of p21, CD95, and other proteins. The p21 protein promotes cell-cycle arrest, whereas the CD95 protein promotes apoptosis. Which of the following will most likely result from a loss of p53 function? A. Rapid cell growth without cell division B. Immediate activation of apoptosis pathways C. Uncontrolled cell proliferation D. Increased expression of p53p53 target genes

C. Uncontrolled cell proliferation

A nucleotide consists of: A.a nitrogen base B.a nitrogen base and a sugar C. a nitrogen base, sugar, and phosphate D. two nitrogen bases, a sugar, and a phosphate

C. a nitrogen base, sugar, and phosphate

Translation begins: A. at the replication fork B. on the lagging strand C. at the start codon D. in the nucleus

C. at the start codon

The twisted ladder shape of DNA is called a: A. hydrogen twist B. deoxyribose flip C. double helix D. double membrane

C. double helix

A stretch of chromosome that codes for a trait can be called a(n): A. chromatid B. replication fork C. gene D. base-pair

C. gene

The enzymes that break hydrogen bonds and unwind DNA are: A. primers B. forks C. helicases D. polymerases

C. helicase

Most cells spend their lives in: A. prophase B. metaphase C. interphase D. telophase

C. interphase

DNA is called the "blueprint of life" because: A. it is like a fingerprint B. it has a blue color C. it contains the plans for building an organism D. it can relay messages to other molecules

C. it contains the plans for building an organism

Okazaki fragments occur on the ___ and are bonded together by ______ A. leading strand, polymerase B. mRNA, anticodons C. lagging strand, ligase D. tRNA, polymerase

C. lagging strand, ligase

Which of the following is NOT a necessary component of translation? A. anticodon B. mRNA C. ligase D. amino acid

C. ligase

Chromosomes line up on the equator during: A. anaphase B. telophase C. metaphase D. interphase

C. metaphase

In the ladder anology of the DNA molecule, the "rungs" of the ladder are: A. sugars B. phosphates C. paired nitrogenous bases D. joined sugar and phosphate

C. paired nitrogenous bases

A spindle forms during which phase? A. G2 B. interphase C. prophase D. metaphase

C. prophase

The outside of the DNA ladder (the legs) is composed of alternating ___ and phosphates. A. adenines B. hydrogens C. sugars D. bases

C. sugars

Adenine always pairs with: A. guanine C. cytosine C. thymine D. another adenine

C. thymine

The process by which RNA is made from DNA: A. synthesis C. translation C. transcription D. replication

C. transcription

D What is the recombination frequency between the genes for petal shape and pollen shape?

The process by which DNA makes a copy of itself is called: A. transcription B. translation C. synthesis D. replication

D,. replication

Cycloheximide (CHX) is a eukaryote protein synthesis inhibitor. It is used in biomedical research to inhibit protein synthesis in eukaryotic cells studied in vitro. Its effects are rapidly reversed by simply removing it from the culture medium. In a translation experiment using a fungus culture, radiolabeled amino acids were added to the culture, allowing the researchers to measure the growth of a single polypeptide chain by measuring counts per minute (CPM). As the chain grew, the CPM increased. After a certain amount of time, CHX was added to the mixture, and the experiment continued. After an additional amount of time, the CHX was removed from the culture medium. Which of the following graphs best predicts the data collected during the experiment? A B C D

Proteins contain ____ different amino acids, whereas DNA and RNA are composed of ___ different nucleotides A. 20, 64 B. 3, 20 C.4, 20 D. 20, 4

D. 20, 4

D. A poly‑A tail is added to RNA.

Different polysaccharides are used by plants for energy storage and structural support. The molecular structures for two common polysaccharides are shown in Figure 1. Starch is used by plants for energy storage, and cellulose provides structural support for cell walls. The monomer used to construct both molecules is glucose. A study determined the effect of two different digestive enzymes, A and B, on these two polysaccharides. Table 1 presents the data from the study. Mammals do not produce digestive enzyme B. However, sheep and cattle are two types of mammals that contain microorganisms in their digestive tract that produce enzyme B. Which of the following would most likely occur if cattle lost the ability to maintain a colony of microorganisms in their digestive tract? A. Cattle would no longer be able to synthesize cellulose. B. Cattle would have to convert cellulose to starch before digesting it. C. Cattle would have to start producing enzyme BB without the help of the bacteria. D. Cattle would no longer be able to use cellulose as a primary source of glucose.

D. Cattle would no longer be able to use cellulose as a primary source of glucose.

Once transcription has been completed, which of the following is NOT necessary for protein synthesis to occur? A. tRNA B. ribosomes C. mRNA D. DNA

D. DNA

Labeled nucleotides were supplied to a cell culture before the cells began DNA replication. A simplified representation of the process for a short segment of DNA is shown in Figure 1. Labeled DNA bases are indicated with an asterisk (*). Figure 1. A simplified representation of the DNA replication process Which of the following best helps explain how the process represented in Figure 1 produces DNA molecules that are hybrids of the original and the newly synthesized strands? A. Each template strand is broken down into nucleotides, which are then used to synthesize both strands of a new DNA molecule. B. Each template strand is broken into multiple fragments, which are randomly assembled into two different DNA molecules. C. Each newly synthesized strand is associated with another newly synthesized strand to form a new DNA molecule. D. Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA molecule.

D. Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA molecule.

A simplified model of a DNA replication fork is represented in Figure 1. The protein labeled Enzyme 1 carries out a specific role in the DNA replication process. Figure 1. A simplified model of a DNA replication fork Which of the following statements best explains the role of Enzyme 1 in the DNA replication process? AEnzyme 1 is a DNA ligase that joins together the DNA fragments at a replication fork to form continuous strands. B. Enzyme 1 is a DNA primase that catalyzes the synthesis of RNA primers on the lagging strand of a replication fork. C. Enzyme 1 is a DNA polymerase that synthesizes new DNA by using the leading and lagging strands of a replication fork as templates. D. Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork.

D. Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork.

Blood clots are formed by a positive feedback loop. Two pathways exist, the extrinsic and intrinsic pathways, which converge during clot formation. There are many clotting factors involved, most of which are proteins. Vitamin K is required for the formation of the active form of several of the clotting factors, including Factor X. A simplified model of the blood clotting process is shown in Figure 1. Figure 1. Simplified model of clotting cascade Warfarin is a drug used to treat certain blood clots. Warfarin blocks the formation of the active form of vitamin K-dependent clotting factors. Based on the model, which of the following best predicts the effects of warfarin on a patient? A. Fibrinogen will form fibrin, but the clot will not form because Factor XIIIXIII will not be synthesized. B. The intrinsic pathway will take over because the clotting factors are part of that pathway. C. Thrombin will be converted to prothrombin because Factor XX will reverse the reaction. D. Factor XX will not be activated, which will prevent thrombin from forming.

D. Factor XX will not be activated, which will prevent thrombin from forming.

Water molecules are polar covalent molecules. There is a partial negative charge near the oxygen atom and partial positive charges near the hydrogen atoms due to the uneven distribution of electrons between the atoms, which results in the formation of hydrogen bonds between water molecules. The polarity of water molecules contributes to many properties of water that are important for biological processes. Which of the following models best demonstrates the arrangement of hydrogen bonds between adjacent water molecules? A. H-O-H--H-O-H B. HH__O--O--HH C. H-O-H--H-O-H-- D. H-O-H--O--HH

D. H-O-H--O--HH

A model showing the cells in anaphase I and anaphase II of meiosis during a nondisjunction event is shown in Figure 1. Figure 1. Model of a nondisjunction event Which of the following best predicts the effect of the chromosomal segregation error shown in Figure 1? A. All of the resulting gametes will have an extra chromosome. B. All of the resulting gametes will be missing a chromosome. C. Half of the resulting gametes will have an extra chromosome and the other half will be missing a chromosome. D. Half of the resulting gametes will have the correct number of chromosomes, and the other half will have an incorrect number of chromosomes.

D. Half of the resulting gametes will have the correct number of chromosomes, and the other half will have an incorrect number of chromosomes.

In anaphase I of meiosis, cohesion between the centromeres of sister chromatids is maintained while homologous chromosomes migrate to opposite poles of the cell along the meiotic spindle as represented in Figure 1. Figure 1. Migration of homologous chromosomes during anaphase I of meiosis A compound that prevents the separation of the homologous chromosomes in anaphase I is being studied. Which of the following questions can be best answered during this study? A. Will the cells produced at the end of meiosis still be genetically identical to each other in the presence of this compound? B. Will the long-term development of the individual be affected by this meiotic error? C. When do the centrosomes start to move apart during meiosis II as compared to meiosis II? D. Is there a pattern to the movement of homologous chromosomes in the presence of this compound?

D. Is there a pattern to the movement of homologous chromosomes in the presence of this compound?

Figure 1 shows a short segment of a double-stranded nucleic acid molecule. Figure 1. A short segment of a double-stranded nucleic acid molecule Which of the following statements is correct about the molecule shown in Figure 1 ? A. It is RNA because of the relative direction of the two strands. B. It is RNA because of the number of different nucleotides found in the molecule. C. It is DNA because of the nature of the hydrogen bonds between guanine and cytosine. D. It is DNA because of the nucleotides present.

D. It is DNA because of the nucleotides present.

Researchers grew seedlings of corn, Zea mays, in loose and compact sand. The researchers measured the amount of time required for the cells in the growing root tips of the seedlings to double in number. The mean cell doubling times for the two groups of seedlings are shown in Figure 1. Figure 1. Mean cell doubling times for the growing root tips of Zea mays seedlings planted in loose or compact sand Based on the sample means, which of the following conclusions about the cells in the growing root tips of Zea mays seedlings is best supported by the results of the experiment? A. The cells of the root tips grow to larger sizes when the seedlings are planted in compact sand than when the seedlings are planted in loose sand. B. The average rate of mitotic cell division is greater for the root tips growing in loose sand than for the root tips growing in compact sand. C. The average cell cycle time is greater for the root tips growing in compact sand than for the root tips growing in loose sand. D. More cells are produced per unit of time in the root tips growing in compact sand than in the root tips growing in loose sand.

D. More cells are produced per unit of time in the root tips growing in compact sand than in the root tips growing in loose sand.

Which of the following best explains how the pattern of DNA arrangement in chromosomes could be used, in most cases, to determine if an organism was a prokaryote or a eukaryote? A. Prokaryotic DNA Eukaryotic DNA Single circular chromosome Multiple circular chromosomes B. Prokaryotic DNA Eukaryotic DNA Multiple chromosomesSingle chromosome C. Prokaryotic DNA Eukaryotic DNA Single linear chromosomeMultiple linear chromosomes D. Prokaryotic DNA Eukaryotic DNA Single circular chromosomeMultiple linear chromosome

D. Prokaryotic DNA Eukaryotic DNA Single circular chromosomeMultiple linear chromosomes Answer D

A cell culture commonly used in research was selected to study the effect of a specific virus on the timing of cell cycle phases. Two separate cultures were started, one untreated and one inoculated with the virus. Both cultures were incubated under identical conditions. After a period of time, 200 cells from each culture were observed and classified as shown in Table 1. Table 1. Number of normal and infected cells found in three phases of the cell cycle Which of the following most accurately describes an observation and an effect of the viral infection indicated by the data in Table 1? A. Normal cells spend 98 percent of their time cycling in and out of interphase. The virus reduces this to 5 percent of the time. B. Twenty percent of the virus-infected cells are in interphase. These cells are no longer part of the cell cycle. C. Forty percent of the virus-infected cells are in interphase. These cells are preparing for replication of genetic material. D. Seventy-five percent of the virus-infected cells are found in mitosis. The virus stimulates frequent cell division.

D. Seventy-five percent of the virus-infected cells are found in mitosis. The virus stimulates frequent cell division.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Based on the information provided in Figure 1 and Figure 2, which of the following best predicts the effects of a mutation in the promoter of the TYR gene that prevents it from being transcribed? A. DNA damage due to UV radiation will be strongly inhibited, resulting in a positive selection pressure. B. DNA damage due to UV radiation will be strongly inhibited, resulting in a negative selection pressure. C. Skin pigmentation will not be able to change, resulting in a positive selection pressure. D. Skin pigmentation will not be able to change, resulting in a negative selection pressure.

D. Skin pigmentation will not be able to change, resulting in a negative selection pressure.

A polypeptide is polymer of amino acids held together by peptide bonds. The process of dehydration synthesis creates these peptide bonds, as shown in Figure 1. Figure 1. Amino acids are linked through the formation of peptide bonds. As shown in Figure 1, an amino acid must have which of the following properties in order to be incorporated into a polypeptide? A. The ability to remain stable in the presence of water molecules B. An R-group that is compatible with the R-group of the last amino acid incorporated C. A central carbon atom that reacts with a nitrogen atom to form the peptide bond D. The ability to form a covalent bond with both its NH2 group and its COOH group

D. The ability to form a covalent bond with both its NH2 group and its COOH group

n corn plants, purple kernel color is dominant to yellow kernel color, and smooth kernels are dominant to wrinkled kernels. Researchers carried out a genetics experiment to investigate whether the genes controlling kernel color and kernel texture segregate independently. In their experiment, the researchers crossed two corn plants that were each heterozygous for both kernel color and kernel texture. The results of the experiment are presented in Table 1. Table 1. Results of a genetic cross between two heterozygous corn plants Using a significance level of p=0.05, which of the following statements best completes a chi-square goodness-of-fit test for a model of independent assortment? A. The calculated chi-square value is 0.66, and the critical value is 0.05. The null hypothesis can be rejected. B. The calculated chi-square value is 0.66, and the critical value is 3.84. The null hypothesis cannot be rejected. C. The calculated chi-square value is 3.91, and the critical value is 5.99. The null hypothesis can be rejected. D. The calculated chi-square value is 3.91, and the critical value is 7.82. The null hypothesis cannot be rejected.

D. The calculated chi-square value is 3.91, and the critical value is 7.82. The null hypothesis cannot be rejected.

Rubber rabbitbrush plants display heritable variation in plant height and insect-induced gall formation. In a field study, researchers investigated the relationship between plant height and gall number for the rubber rabbitbrush plants in a natural population. The researchers used the data they collected to perform a chi-square test of independence. The null hypothesis for the chi-square test was that plant height and gall number are independent. The data for the chi-square test are presented in Table 1. Table 1. Data for the chi-square test of independence The researchers calculated a chi-square value of 29.25. If there are three degrees of freedom and the significance level is p=0.05, which of the following statements best completes the chi-square test? A. The critical value is 0.05, and the null hypothesis cannot be rejected because the calculated chi-square value is greater than the critical value. B. The critical value is 0.05, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value. C. The critical value is 7.82, and the null hypothesis cannot be rejected because the calculated chi-square value is greater than the critical value. D. The critical value is 7.82, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value.

D. The critical value is 7.82, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value.

Himalayan rabbits are a breed of rabbits with highly variable fur color. If genetically similar rabbits are raised in environments that have different temperature conditions, the rabbits can have different color patterns. Which of the following statements best explains how the fur color can be different in Himalayan rabbits raised under different temperature conditions? A. The genotype does not contribute to coat color in Himalayan rabbits. B. The phenotype determines the genotype of coat color in Himalayan rabbits. C. Different environments cause specific mutations in the genes controlling pigment production. D. The environment determines how the genotype is expressed.

D. The environment determines how the genotype is expressed.

Which feature of model 1 best illustrates how biological information is coded in a DNA molecule? A. The 5′5′ and 3′3′ labels at the ends of each strand B. The labeling of the hydrogen bonds between base pairs C. The lines connecting sugars and phosphate groups that represent covalent bonds D. The linear sequence of the base pairs

D. The linear sequence of the base pairs

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Researchers discovered a mutant form of the TYR gene with a deletion of a single guanine nucleotide in the beginning of the coding sequence. Which of the following best predicts the phenotype of an individual who is homozygous for this TYR mutation? A. The mutation will cause a single amino acid change in the TYR protein, which will not be enough to disrupt its function. Therefore, those with this mutation will produce melanin in the hair, skin, and eyes and tan in response to UV radiation. B. The mutation will cause a single amino acid change in the TYR protein, leading to a nonfunctional TYR protein. Therefore, those with this mutation will lack melanin in the hair, skin, and eyes and will not tan in response to UV radiation. C. The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Since the TRP1 and TRP2 genes were not affected, the TRP1 and TRP2 proteins will fill the role of the TYR protein. Therefore, those with this mutation will produce melanin in the hair, skin, and eyes in response to UV radiation. D. The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Individuals with this mutation will lack melanin in their hair, skin, and eyes and will not tan in response to UV radiation.

D. The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Individuals with this mutation will lack melanin in their hair, skin, and eyes and will not tan in response to UV radiation.

The beta-2 adrenergic receptor is a membrane-bound protein that regulates several cellular processes, including the synthesis and breakdown of glycogen. The receptor binds specifically to the hormone epinephrine. The binding of epinephrine to the beta-2 adrenergic receptor triggers a signal transduction cascade that controls glycogen synthesis and breakdown in the cell. A simplified model of the signal transduction cascade is represented in Figure 1. Figure 1. A simplified model of the signal transduction cascade triggered by epinephrine binding to the beta-2 adrenergic receptor Which of the following outcomes will most likely result from the inactivation of the beta-2 adrenergic receptor? A. The cellular concentration of cyclic AMP will increase. B. The enzymatic activity of protein kinase A will increase. C. The activation of glycogen phosphorylase will increase. D. The rate of glycogen synthesis in the cell will increase.

D. The rate of glycogen synthesis in the cell will increase.

Exposure to ultraviolet (UV) radiation is the leading cause of skin cancer in humans. Figure 1 shows a model of how UV exposure damages DNA. Figure 1. Model of damage to DNA caused by UV exposure Which of the following statements best explains what is shown in Figure 1 ? A. UV exposure triggers DNA replication, which results in rapid cell proliferation. B. Naturally occurring dimers are removed by the UV photons, causing misshapen DNA, which results in replication errors. C. The hydrogen bonds between base pairs absorb the UV photons, separating the two DNA strands, which results in rapid DNA replication. D. UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors.

D. UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors.

Small single-stranded RNA molecules called microRNAs (miRNAs) are capable of base pairing with specific binding sites in the 3′ untranslated region of many mRNA transcripts. Transcription of gene Q yields an mRNA transcript that contains such an miRNA binding site, which can associate with miRNA‑delta, a specific miRNA molecule. Which of the following best supports the claim that binding of miRNA‑delta to the miRNA binding site inhibits translation of gene Q mRNA? A. When the promoter for gene Q is altered, transcription is inhibited. B. Translation of Q mRNA is inhibited regardless of whether the miRNA binding site sequence is altered. C. Translation of Q mRNA is inhibited in the absence of miRNA‑delta. D. When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta

D. When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta.

The trp operon in E. coli is an example of a repressible operon that consists of genes coding for enzymes used to synthesize tryptophan. When tryptophan levels are high, the operon is turned off and these genes are not transcribed. However, it is also known that tryptophan does not bind directly to the operator DNA sequence. A regulatory gene called trpR has also been discovered although it is not part of the trp operon. The proposed model of how tryptophan acts as a corepressor is shown in Figure 1. Figure 1. Model of proposed regulation of the trp operon by corepressors trp repressor and tryptophan Which of the following evidence best supports a claim that tryptophan functions as a corepressor? A. Normal expression of trpR causes the trp operon to be transcribed regardless of tryptophan levels. B. When the operator sequence is mutated, the trp operon is not transcribed. C. The trpR gene codes for a repressor protein that has a DNA binding domain. D. When trpR is mutated, the trp operon is transcribed regardless of tryptophan levels.

D. When trpR is mutated, the trp operon is transcribed regardless of tryptophan levels.

A nucleotide consists of: A. a phosphate and a base B. a phosphate, and a sugar C. a base and an amino acid D. a phosphate, a sugar, and a base

D. a phosphate, a sugar, and a base

Chromosomes are paired together based on their: A. size B. banding pattern C. centromere location D. all of these

D. all of these

RNA differs from DNA in that: A. it has a different kind of sugar B. it is single stranded C. it has uracil D. all of these

D. all of these

Which of the following can be determined from a karyotype? A. the sex of the individual B. whether the individual has Down Syndrome C. The number of chromosomes present D. all of these

D. all of these

Which of the following is required for DNA replication to occur? A.DNA helicase B. DNA ligase C.DNA polymerase D. all of these

D. all of these

The "rungs" of the DNA ladder are made of: A. phosphates and hydrogen B. glucose and sugars C. sugars and phosphates D. base pairs

D. base pairs

The "rungs" of the DNA ladder are made of: A. phosphates and bases B. sugars C. sugars and phosphates D. bases

D. bases

G proteins are a family of receptor proteins that are involved in transmitting signals from outside a cell to inside a cell. When a signaling molecule binds to a G protein, the G protein is activated. The G protein then activates an enzyme that produces a second messenger called cAMP. Which of the following describes a critical role of cAMP during the transduction stage of a G protein signal transduction pathway? A. cAMP carries the signal to the nucleus of the cell and results in new sequences of nucleotides being added to the cell's DNA. B. cAMP binds the extracellular signal molecule and carries it to the intracellular target specified by the signal. C. cAMP modifies a specific monomer so that it can be added to an elongating structural macromolecule. D. cAMP results in the activation of an enzyme that amplifies the signal by acting on many substrate molecules.

D. cAMP results in the activation of an enzyme that amplifies the signal by acting on many substrate molecules.

The twisted ladder shape of the DNA molecule is known as the: A. isohedron B. deoxyribose twist C. hydrogren twist D. double helix

D. duoble helix

The DNA molecule is held together by: A. magnetism B. glucose C. glue D. hydrogen bonds

D. hydrogen bonds

Which of the following is NOT part of mitosis A. prophase B. metaphase C. telophase D. interphase

D. interphase

Which of the following takes the genetic code to the cytoplasm: A. DNA B. deoxyribose C. tRNA D. mRNA

D. mRNA

During replication, what enzyme adds complimentary bases? A. helicase B. synthesase C. replicase D. polymerase

D. polymerase

Which of the following is NOT part of the chromosome? A. kinetochore B. chromatid C. centromere D. spindle

D. spindle

The two men who established the structure of DNA are: A. Berkeley and Fry B. Einstein and Gilbert C. Darwin and Lamarke D. Watson and Crick

D. watson and crick

Which of the following best explains a distinction between metaphase I and metaphase II? A. The nuclear membrane breaks down during metaphase I but not during metaphase II. B. Chromosomes align at the equator of the cell during metaphase II but not during metaphase I. C. The meiotic spindle is needed during metaphase I but not during metaphase II. D. Homologous pairs of chromosomes are aligned during metaphase I, but individual chromosomes are aligned during metaphase II.

DHomologous pairs of chromosomes are aligned during metaphase I, but individual chromosomes are aligned during metaphase II.

What is the long name for DNA? A. deoxynucleic acid B. deoxyribonucleic acid C. denatured ribonucleic acid D. deoxoribonuclear acid

b. deoxyribonucleic acid

Meiosis is a type of cell division that produces: A. zygotes B. chromosomes C. DNA D. gametes

gametes

Which of the following distinguishes prophase 1 of meiosis from prophase of mitosis? A. homologous chromosomes pair up B. spindle forms C.nuclear membrane breaks down D. chromosomes become visible

homologous chromosomes pair up

Crossing-over occurs during: A. anaphase 1 B. metaphase 1 C. prophase 1 D. prophase 2

ap bio midterm

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