Most repressor proteins are allosteric. Which of the following binds with the repressor to alter its conformation?
A) inducer
B) promoter
C) RNA polymerase
D) transcription factor
E) cAMP

Answer: A

A mutation that inactivates the regulatory gene of a repressible operon in an E. coli cell would result in
A) continuous transcription of the structural gene controlled by that regulator.
B) complete inhibition of transcription of the structural gene controlled by that regulator.
C) irreversible binding of the repressor to the operator.
D) inactivation of RNA polymerase by alteration of its active site.
E) continuous translation of the mRNA because of alteration of its structure.

Answer: A

The lactose operon is likely to be transcribed when
A) there is more glucose in the cell than lactose.
B) the cyclic AMP levels are low.
C) there is glucose but no lactose in the cell.
D) the cyclic AMP and lactose levels are both high within the cell.
E) the cAMP level is high and the lactose level is low.

Answer: D

Transcription of the structural genes in an inducible operon
A) occurs continuously in the cell.
B) starts when the pathway's substrate is present.
C) starts when the pathway's product is present.
D) stops when the pathway's product is present.
E) does not result in the production of enzymes.

Answer: B

For a repressible operon to be transcribed, which of the following must occur?
A) A corepressor must be present.
B) RNA polymerase and the active repressor must be present.
C) RNA polymerase must bind to the promoter, and the repressor must be inactive.
D) RNA polymerase cannot be present, and the repressor must be inactive.
E) RNA polymerase must not occupy the promoter, and the repressor must be inactive.

Answer: C

Allolactose, an isomer of lactose, is formed in small amounts from lactose. An E. coli cell is presented for the first time with the sugar lactose (containing allolactose) as a potential food source. Which of the following occurs when the lactose enters the cell?
A) The repressor protein attaches to the regulator.
B) Allolactose binds to the repressor protein.
C) Allolactose binds to the regulator gene.
D) The repressor protein and allolactose bind to RNA polymerase.
E) RNA polymerase attaches to the regulator.

Answer: B

Altering patterns of gene expression in prokaryotes would most likely serve the organism's survival in which of the following ways?
A) organizing gene expression so that genes are expressed in a given order
B) allowing each gene to be expressed an equal number of times
C) allowing the organism to adjust to changes in environmental conditions
D) allowing young organisms to respond differently from more mature organisms
E) allowing environmental changes to alter the prokaryote's genome

Answer: C

In response to chemical signals, prokaryotes can do which of the following?
A) turn off translation of their mRNA
B) alter the level of production of various enzymes
C) increase the number and responsiveness of their ribosomes
D) inactivate their mRNA molecules
E) alter the sequence of amino acids in certain proteins

Answer: B

If glucose is available in the environment of E. coli, the cell responds with a very low concentration of cAMP. When the cAMP increases in concentration, it binds to CAP. Which of the following would you expect to be a measurable effect?
A) decreased concentration of the lac enzymes
B) increased concentration of the trp enzymes
C) decreased binding of the RNA polymerase to sugar metabolism-related promoters
D) decreased concentration of alternative sugars in the cell
E) increased concentrations of sugars such as arabinose in the cell

Answer: E

There is a mutation in the repressor that results in a molecule known as a super-repressor because it represses the lac operon permanently. Which of these would characterize such a mutant?
A) It cannot bind to the operator.
B) It cannot make a functional repressor.
C) It cannot bind to the inducer.
D) It makes molecules that bind to one another.
E) It makes a repressor that binds CAP.

Answer: C

Which of the following mechanisms is (are) used to coordinate the expression of multiple, related genes in eukaryotic cells?
A) Genes are organized into clusters, with local chromatin structures influencing the expression of all the genes at once.
B) The genes share a common intragenic sequence, and allow several activators to turn on their transcription, regardless of location.
C) The genes are organized into large operons, allowing them to be transcribed as a single unit.
D) A single repressor is able to turn off several related genes.
E) Environmental signals enter the cell and bind directly to promoters.

Answer: A

Genomic imprinting, DNA methylation, and histone acetylation are all examples of
A) genetic mutation.
B) chromosomal rearrangements.
C) karyotypes.
D) epigenetic phenomena.
E) translocation.

Answer: D

Two potential devices that eukaryotic cells use to regulate transcription are
A) DNA methylation and histone amplification.
B) DNA amplification and histone methylation.
C) DNA acetylation and methylation.
D) DNA methylation and histone modification.
E) histone amplification and DNA acetylation.

Answer: D

During DNA replication,
A) all methylation of the DNA is lost at the first round of replication.
B) DNA polymerase is blocked by methyl groups, and methylated regions of the genome are therefore left uncopied.
C) methylation of the DNA is maintained because methylation enzymes act at DNA sites where one strand is already methylated and thus correctly methylates daughter strands after replication.
D) methylation of the DNA is maintained because DNA polymerase directly incorporates methylated nucleotides into the new strand opposite any methylated nucleotides in the template.
E) methylated DNA is copied in the cytoplasm, and unmethylated DNA is copied in the nucleus.

Answer: C

Transcription factors in eukaryotes usually have DNA binding domains as well as other domains that are also specific for binding. In general, which of the following would you expect many of them to be able to bind?
A) repressors
B) ATP
C) protein-based hormones
D) other transcription factors
E) tRNA

Answer: D

Gene expression might be altered at the level of post-transcriptional processing in eukaryotes rather than prokaryotes because of which of the following?
A) Eukaryotic mRNAs get 5' caps and 3' tails.
B) Prokaryotic genes are expressed as mRNA, which is more stable in the cell.
C) Eukaryotic exons may be spliced in alternative patterns.
D) Prokaryotes use ribosomes of different structure and size.
E) Eukaryotic coded polypeptides often require cleaving of signal sequences before localization.

Answer: C

Among the newly discovered small noncoding RNAs, one type reestablishes methylation patterns during gamete formation and block expression of some transposons. These are known as
A) miRNA.
B) piRNA.
C) snRNA.
D) siRNA.
E) RNAi.

Answer: B

The fact that plants can be cloned from somatic cells demonstrates that
A) differentiated cells retain all the genes of the zygote.
B) genes are lost during differentiation.
C) the differentiated state is normally very unstable.
D) differentiated cells contain masked mRNA.
E) differentiation does not occur in plants.

Answer: A

In animals, embryonic stem cells differ from adult stem cells in that
A) embryonic stem cells are totipotent, and adult stem cells are pluripotent.
B) embryonic stem cells are pluripotent, and adult stem cells are totipotent.
C) embryonic stem cells have more genes than adult stem cells.
D) embryonic stem cells have fewer genes than adult stem cells.
E) embryonic stem cells are localized to specific sites within the embryo, whereas adult stem cells are spread throughout the body.

Answer: A

What is considered to be the first evidence of differentiation in the cells of an embryo?
A) cell division occurring after fertilization
B) the occurrence of mRNAs for the production of tissue-specific proteins
C) determination of specific cells for certain functions
D) changes in the size and shape of the cell
E) changes resulting from induction

Answer: B

Your brother has just purchased a new plastic model airplane. He places all the parts on the table in approximately the positions in which they will be located when the model is complete. His actions are analogous to which process in development?
A) morphogenesis
B) determination
C) induction
D) differentiation
E) pattern formation

Answer: E

If a Drosophila female has a homozygous mutation for a maternal effect gene,
A) she will not develop past the early embryonic stage.
B) all of her offspring will show the mutant phenotype, regardless of their genotype.
C) only her male offspring will show the mutant phenotype.
D) her offspring will show the mutant phenotype only if they are also homozygous for the mutation.
E) only her female offspring will show the mutant phenotype.

Answer: B

Cell differentiation always involves
A) the production of tissue-specific proteins, such as muscle actin.
B) the movement of cells.
C) the transcription of the myoD gene.
D) the selective loss of certain genes from the genome.
E) the cell's sensitivity to environmental cues, such as light or heat.

Answer: A

Most human-infecting viruses are maintained in the human population only. However, a zoonosis is a disease that is transmitted from other vertebrates to humans, at least sporadically, without requiring viral mutation. Which of the following is the best example of a zoonosis?
A) rabies
B) herpesvirus
C) smallpox
D) HIV
E) hepatitis virus

Answer: A

Which of the following is characteristic of the lytic cycle?
A) Many bacterial cells containing viral DNA are produced.
B) Viral DNA is incorporated into the host genome.
C) The viral genome replicates without destroying the host.
D) A large number of phages are released at a time.
E) The virus-host relationship usually lasts for generations.

Answer: D

Which of the following statements describes the lysogenic cycle of lambda (λ) phage?
A) After infection, the viral genes immediately turn the host cell into a lambda-producing factory, and the host cell then lyses.
B) Most of the prophage genes are activated by the product of a particular prophage gene.
C) The phage genome replicates along with the host genome.
D) Certain environmental triggers can cause the phage to exit the host genome, switching from the lytic to the lysogenic.
E) The phage DNA is incorporated by crossing over into any nonspecific site on the host cell's DNA.

Answer: C

Why do RNA viruses appear to have higher rates of mutation?
A) RNA nucleotides are more unstable than DNA nucleotides.
B) Replication of their genomes does not involve proofreading.
C) RNA viruses replicate faster.
D) RNA viruses can incorporate a variety of nonstandard bases.
E) RNA viruses are more sensitive to mutagens.

Answer: B

Which viruses have single-stranded RNA that acts as a template for DNA synthesis?
A) lytic phages
B) proviruses
C) viroids
D) bacteriophages
E) retroviruses

Answer: E

Which of the following describes plant virus infections?
A) They can be controlled by the use of antibiotics.
B) They are spread via the plasmodesmata.
C) They have little effect on plant growth.
D) They are seldom spread by insects.
E) They can never be passed vertically.

Answer: B

RNA viruses require their own supply of certain enzymes because
A) host cells rapidly destroy the viruses.
B) host cells lack enzymes that can replicate the viral genome.
C) these enzymes translate viral mRNA into proteins.
D) these enzymes penetrate host cell membranes.
E) these enzymes cannot be made in host cells.

Answer: B

Although not present in all bacteria, this cell covering often enables cells that possess it to resist the defenses of host organisms, especially their phagocytic cells.
A) endospore
B) sex pilus
C) cell wall
D) capsule

Answer: D

Prokaryotes' essential genetic information is located in the
A) nucleolus.
B) nucleoid.
C) nucleosome.
D) plasmids.
E) exospore.

Answer: B

Which of the following is an important source of endotoxin in gram-negative species?
A) endospore
B) sex pilus
C) flagellum
D) cell wall
E) capsule

Answer: D

Which of the following is least associated with the others?
A) horizontal gene transfer
B) genetic recombination
C) conjugation
D) transformation
E) binary fission

Answer: E

In Fred Griffith's experiments, harmless R strain pneumococcus became lethal S strain pneumococcus as the result of which of the following?
1. horizontal gene transfer
2. transduction
3. conjugation
4. transformation
5. genetic recombination

A) 2 only
B) 4 only
C) 2 and 5
D) 1, 3, and 5
E) 1, 4, and 5

Answer: E

Which statement about the domain Archaea is true?
A) Genetic prospecting has recently revealed the existence of many previously unknown archaean species.
B) No archaeans can reduce CO₂ to methane.
C) The genomes of archaeans are unique, containing no genes that originated within bacteria.
D) No archaeans can inhabit solutions that are nearly 30% salt.
E) No archaeans are adapted to waters with temperatures above the boiling point.

Answer: A

Genetic variation in bacterial populations cannot result from
A) transduction.
B) transformation
C) conjugation
D) mutation.
E) meiosis.

Answer: E

Which of the following statements is not true?
A) Archaea and bacteria have different membrane lipids.
B) Both archaea and bacteria generally lack membrane-enclosed organelles.
C) The cell walls of archaea lack peptidoglycan.
D) Only bacteria have histones associated with DNA.
E) Only some archaea use CO₂ to oxidize H₂, releasing methan

Answer: D

Which of the following is (are) unique to animals?
A) cells that have mitochondria
B) the structural carbohydrate, chitin
C) nervous conduction and muscular movement
D) heterotrophy
E) Two of these responses are correct.

Answer: C

An adult animal that possesses bilateral symmetry is most certainly also
A) triploblastic.
B) a deuterostome.
C) eucoelomate.
D) highly cephalized.

Answer: A

Soon after the coelom begins to form, a researcher injects a dye into the coelom of a deuterostome embryo. Initially, the dye should be able to flow directly into the
A) blastopore.
B) blastocoel.
C) archenteron.
D) pseudocoelom.

Answer: C

What distinguishes a coelomate animal from a pseudocoelomate animal is that coelomates
A) have a body cavity, whereas pseudocoelomates have a solid body.
B) contain tissues derived from mesoderm, whereas pseudocoelomates have no such tissue.
C) have a body cavity completely lined by mesodermal tissue, whereas pseudocoelomates do not.
D) have a complete digestive system with mouth and anus, whereas pseudocoelomates have a digestive tract with only one opening.
E) have a gut that lacks suspension within the body cavity, whereas pseudocoelomates have mesenteries that hold the digestive system in place.

Answer: C

The blastopore is a structure that first becomes evident during
A) fertilization.
B) gastrulation.
C) the eight-cell stage of the embryo.
D) coelom formation.
E) cleavage.

Answer: B

Which of the following is descriptive of protostomes?
A) spiral and indeterminate cleavage, blastopore becomes mouth
B) spiral and determinate cleavage, blastopore becomes mouth
C) spiral and determinate cleavage, blastopore becomes anus
D) radial and determinate cleavage, blastopore becomes anus
E) radial and determinate cleavage, blastopore becomes mouth

Answer: B

Which of the following characteristics generally applies to protostome development?
A) radial cleavage
B) determinate cleavage
C) diploblastic embryo
D) blastopore becomes the anus
E) archenteron absent

Answer: B

Protostome characteristics generally include which of the following?
A) a mouth that develops secondarily, and far away from the blastopore
B) radial body symmetry
C) radial cleavage
D) determinate cleavage
E) absence of a body cavity

Answer: D

A student encounters an animal embryo at the eight-cell stage. The four smaller cells that comprise one hemisphere of the embryo seem to be rotated 45 degrees and to lie in the grooves between larger, underlying cells (i.e., spiral cleavage).

If we were to separate these eight cells and attempt to culture them individually, then what is most likely to happen?
A) All eight cells will die immediately.
B) Each cell may continue development, but only into a nonviable embryo that lacks many parts.
C) Each cell may develop into a full-sized, normal embryo.
D) Each cell may develop into a smaller-than-average, but otherwise normal, embryo.

Answer: B

A student encounters an animal embryo at the eight-cell stage. The four smaller cells that comprise one hemisphere of the embryo seem to be rotated 45 degrees and to lie in the grooves between larger, underlying cells (i.e., spiral cleavage).

This embryo may potentially develop into a(n)
A) turtle.
B) earthworm.
C) sea star.
D) fish.
E) sea urchin.

Answer: B

Among the characteristics unique to animals is
A) gastrulation.
B) multicellularity.
C) sexual reproduction.
D) flagellated sperm.
E) heterotrophic nutrition

Answer:A