Test 2 Eoc Questions

Front 1

The water necessary for photosynthesis
a. is split into H2 and O2.
b. is directly involved in the synthesis of carbohydrate.
c. provides the electrons to replace lost electrons in photosystem II.
d. provides H+ needed to synthesize G3P.
e. none of the above.

Back 1

c. The water is split to provide electrons needed to replace lost ones in photosystem II.

Front 2

The reaction center pigment differs from the other pigment molecules of the light-harvesting complex in that
a. the reaction center pigment is a carotenoid.
b. the reaction center pigment absorbs light energy and transfer that energy to other molecules without the transfer of electrons.
c. the reaction center pigment transfers excited electrons to other molecules.
d. the other pigments transfer high-energy electrons to other pigment molecules.
e. the reaction center acts as an ATP synthase to produce ATP.

Back 2

Though the pigment molecules of the light-harvesting complex can transfer energy absorbed from light to other molecules, this is done without the transfer of electrons. The reaction center pigment actually transfers the excited electrons to other molecules to capture the energy that is used to do cellular work.

Front 3

The cyclic electron flow that occurs in photosystem I produces
a. NADPH.
b. oxygen.
c. ATP.
d. all of the above.
e. a and c only.

Back 3

Cyclic electron flow of photosystem I produces additional ATP that is necessary for the Calvin cycle.

Front 4

4. During the light reactions, the high-energy electron from P680
a. eventually moves to NADP+.
b. becomes incorporated in water molecules.
c. is pumped into the thylakoid space to drive ATP production.
d. provides the energy necessary to split water molecules.
e. falls back to the low energy state in photosystem II.

Back 4

a. The high-energy electron of P680 eventually is transferred to NADP+.

Front 5

During the first phase of the Calvin cycle, carbon dioxide is incorporated into ribulose bisphosphate by
a. oxaloacetate.
b. rubisco.
c. RuBP.
d. quinone.
e. G3P.

Back 5

b. Rubisco is the enzyme that catalyzes the reaction that incorporates carbon dioxide into RuBP.

Front 6

The NADPH produced during the light reactions is necessary for
a. the carbon fixation phase that incorporates carbon dioxide into an organic
molecule of the Calvin cycle.
b. the reduction phase that produces carbohydrates in the Calvin cycle.
c. the regeneration of RuBP of the Calvin cycle.
d. all of the above.
e. a and b only.

Back 6

b. The NADPH is necessary for the reduction phase of the Calvin cycle.

Front 7

The majority of the G3P produced during the reduction and carbohydrate production phase is used to produce
a. glucose.
b. ATP.
c. RuBP to continue the cycle.
d. rubisco.
e. all of the above.

Back 7

c. Ten of the twelve G3P molecules produced in the second phase of the Calvin cycle is used to regenerate RuBP, allowing the cycle to continue with additional carbon fixation.

Front 8

Photorespiration
a. is the process where plants use sunlight to make ATP.
b. is an inefficient way plants can produce organic molecules but in the process use O2 and release CO2.
c. is a process that plants use to convert light energy to NADPH.
d. occurs in the thylakoid space.
e. is the normal process of carbohydrate production in cool, moist environments.

Back 8

b. Photorespiration is the process where plants produce organic molecules by using O2 and releasing CO2.

Front 9

Photorespiration is avoided in C4 plants because
a. these plants separate the formation of a four-carbon molecule from the rest of the Calvin cycle in different cells.
b. these plants only carry out anaerobic respiration.
c. the enzyme PEP functions to maintain high CO2 concentrations in the bundle-sheath cells.
d. all of the above.
e. a and c only.

Back 9

e. Photorespiration is avoided in C4 plants because carbon fixation occurs in one type of cell and the rest of the Calvin cycle occurs in another type of cell. Also, through the activity of the PEP, CO2 levels are high, reducing the probability of photorespiration.

Front 10

Plants that are commonly found in hot, dry environments that carry out carbon fixation mainly at night are
a. oak trees.
b. C3 plants.
c. CAM plants.
d. all of the above.
e. a and b only.

Back 10

c. CAM plants are a type of C4 plant that carry out carbon fixation at night.

Front 11

The function of the extracellular matrix (ECM) in most multicellular organisms is
a. to provide strength.
b. to provide structural support.
c. to organize cells and other body parts.
d. cell signaling.
e. all of the above.

Back 11

e. The ECM serves all of these functions.

Front 12

The protein found in the ECM of animals that provides strength and resistance to tearing when stretched is
a. elastin.
b. cellulose.
c. collagen.
d. laminin.
e. fibronectin.

Back 12

c. Collagen provides tensile strength to animal tissues.

Front 13

The polysaccharide that forms the hard outer covering of many invertebrates is
a. collagen.
b. chitin.
c. chondroitin sulfate.
d. pectin.
e. cellulose.

Back 13

b. Chitin is a polysaccharide found in the hard outer covering of many invertebrates.

Front 14

The single most abundant organic molecule of Earth is __________ and it is the main macromolecule of the _______________.
a. collagen, connective tissue of animals
b. chitin, muscle tissue of animals
c. cellulose, primary cell wall of plants
d. integrins, cell junctions in plants
e. pectin, secondary cell wall of plants

Back 14

c. Cellulose is the main macromolecule of the primary cell wall of plants and is very likely the most abundant organic molecule on Earth.

Front 15

__________________ are proteins that attach animal cells to the ECM.
a. Cadherins
b. Integrins
c. Occludins
d. Tight junctions
e. Desmosomes

Back 15

b. Integrins, a type of CAM, are proteins that attach animal cells to the ECM.

Front 16

The gap junctions of animal cells differ from the plasmodesmata of plant cells in that
a. gap junctions serve as communicating junctions and plasmodesmata serve as adhesion junctions.
b. gap junctions prevent extracellular material from moving between adjacent cells but the plasmodesmata do not.
c. gap junctions allow for direct exchange of cellular material between cells, but plasmodesmata cannot allow the same type of exchange.
d. gap junctions are formed by specialized proteins that form channels through the membranes of adjacent cells, but plasmodesmata are not formed by specialized proteins.
e. all of the above

Back 16

d. The plasmodesmata of plant cells are channels between cells but the channel is formed because the plasma membrane of one cell is continuous with the plasma membrane of an adjacent cell. Gap junctions of animal cells, however, are formed by specialized proteins that connect the two membranes and create a channel between the two cells.

Front 17

Which of the following is involved in the process of tissue and organ formation in multicellular organisms?
a. cell division
b. cell differentiation
c. cell connections
d. cell growth
e. All of the above

Back 17

e. All of these processes are needed for tissue and organ formation. Migration and apoptosis also are important in tissue and organ formation in animals.

Front 18

The tissue type common to animals that functions in the conduction of electrical signals is
a. epithelial.
b. dermal.
c. muscle.
d. nervous.
e. ground.

Back 18

d. Nervous tissue is specialized for conducting electrical signals.

Front 19

Photosynthesis occurs mainly in the ______________ tissue of plants.
a. vascular
b. dermal
c. parenchyma
d. collenchyma
e. sclerenchyma

Back 19

c. Parenchyma is the primary site for photosynthesis.

Front 20

Which of the following is not a correct statement in the comparison of plant tissues to animal tissues?
a. Nervous tissue of animals plays the same role as vascular tissue in plants.
b. The dermal tissue of plants is similar to epithelial tissue of animals in that both provide a covering for the organism.
c. The epithelial tissue of animals and the dermal tissue of plants have special characteristics that limit the movement of extracellular material between cell layers.
d. The ground tissue of plants and the connective tissue of animals provide structural support for the organism.
e. All of the above are correct comparisons between plant and animal tissues.

Back 20

a. Nervous tissue in animals is involved in cell signaling, while vascular tissue in plants is involved in the transport of materials throughout the plant.

Front 21

Why did researchers initially believe the genetic material was protein?
a. Proteins are more biochemically complex than DNA.
b. Proteins are found only in the nucleus, but DNA is found in many areas of the cell.
c. Proteins are much larger molecules and can store more information than DNA.
d. All of the above.
e. Both a and c.

Back 21

a. Researchers knew that chromosomes were composed of proteins and DNA. Proteins are polymers of amino acids, which there are 20 different types. DNA is a polymer of nucleotides, which there are only 4 types. Researchers felt that the higher complexity of proteins was evidence of its ability to store information.

Front 22

Considering the components of a nucleotide, what component always determines whether the nucleotide would be incorporated into a DNA strand or an RNA strand?
a. phosphate group
b. pentose sugar
c. nitrogenous base
d. both b and c

Back 22

b. The pentose sugar is the component that determines to what type of nucleic acid the nucleotide belongs. Ribose sugar would indicate RNA and deoxyribose sugar would indicate DNA.

Front 23

Which of the following equations would be appropriate when considering DNA base composition?
a. %A + %T = %G + %C
b. %A = %G
c. %A = %G = %T = %C
d. %A + %G = %T + %C

Back 23

d. Chargaff’s rule indicates a relationship between the % purines and % pyrimidines. According to the DNA model proposed by Watson and Crick, base pairs consist of a purine and a pyrimidine. This would indicate that a double-stranded DNA molecule should have equal percentages of purines and pyrimidines.

Front 24

If the sequence of a section of DNA is 5’-CGCAACTAC-3’, what is the appropriate sequence for the opposite strand?
a. 5’-GCGTTGATG-3’
b. 3’-ATACCAGCA-5’
c. 5’-ATACCAGCA-3’
d. 3’-GCGTTGATC-5’

Back 24

d. When determining the correct sequence of the opposite strand, one must consider the proper base pairing and directionality of the strand. Considering the base pairing, adenine base pairs with thymine and guanine base pairs with cytosine. In double-stranded DNA molecules, the two strands are antiparallel, meaning the orientation of one strand is 5’ to 3’, while the orientation of the opposite strand is 3’ to 5’.

Front 25

Of the following statements, which is correct when considering the process of DNA replication?
a. New DNA molecules are composed of two completely new strands
b. New DNA molecules are composed of one strand from the old molecule and one new strand.
c. New DNA molecules are composed of strands that are a mixture of sections from the old molecule and sections that are new.
d. None of the above.

Back 25

b. DNA is replicated semi-conservatively, where each new molecule is composed of an old strand, parental strand, and one new strand, daughter strand.

Front 26

Meselson and Stahl were able to demonstrate semiconservative replication in E. coli by
a. using radioactive isotopes of phosphorus to label the old strand and visually determining the relationship of old and new DNA strands.
b. using different enzymes to eliminate old strands from DNA.
c. using isotopes of nitrogen to label the DNA and determining the relationship of old and new DNA strands by density differences of the new molecules.
d. labeling viral DNA before it was incorporated into a bacterial cell and visually determining the location of the DNA after centrifugation.

Back 26

c. Meselson and Stahl used different isotopes of nitrogen to determine the composition of newly synthesized DNA molecules. By analyzing the density differences over several generations of replication, Meselson and Stahl were able to present data that were consistent with only semi-conservative replication.

Front 27

During replication of a DNA molecule, the daughter strands are not produced in exactly the same manner. One strand, the leading strand, is made toward the replication fork, while the lagging strand is made in fragments in the opposite direction. This difference in the synthesis of the two strands is the result of
a. DNA polymerase is not efficient enough to make two “good” strands of DNA.
b. the two parental strands are antiparallel, and DNA polymerase makes DNA only in the 5' to 3' direction
c. the lagging strand is the result of DNA breakage due to UV light.
d. the cell does not contain enough nucleotides to make two complete strands.

Back 27

b. The two parental strands are antiparallel and DNA polyermase makes DNA only in the 5' to 3' direction. Because DNA polymerase can add new nucleotides only to the 3’ end of the new strands, the two daughter strands are “growing” in opposite directions. The leading strand follows the replication fork and can be made as a continuous strand. The lagging strand grows in the opposite direction, so synthesis has to be initiated many times along the parental strand, producing the Okazaki fragments.

Front 28

Considering the different proteins involved in DNA replication and their functions, which of the following does not represent an accurate relationship?
a. DNA helicase unwinds the DNA molecule.
b. DNA primase produces the Okazaki fragments of the lagging strand.
c. DNA topoisomerase reduces the coiling ahead of the replication fork.
d. Single-strand binding proteins prevent the double helix from re-forming.

Back 28

b. DNA primase produces short RNA primers that allows DNA polymerase to attach DNA nucleotides. DNA polymerase makes most of the Okazaki fragments.

Front 29

Which of the following does not promote the fidelity of DNA replication?
a. DNA polymerase will not form bonds between nucleotides if there is a mismatched base.
b. DNA polymerase has the ability to recognize mismatched bases and remove them.
c. Hydrogen bonds that hold the two strands together are more stable in correctly matched bases than between mismatched bases.
d. All of the above are correct.

Back 29

d. The hydrogen bonding between correct bases increases the stability of the molecule and the DNA polymerase activity increases the fidelity of DNA replication.

Front 30

Most organisms have more than one form of DNA polymerase. Some forms play specific roles in replication, while others play roles in DNA repair. What is the proposed mechanism for the evolution of the multiple forms of DNA polymerase?
a. Increased copies of polymerase genes due to chromosome number changes over several generations.
b. Species hybridization may introduce new forms of polymerases.
c. Gene duplication followed by mutations that alter the activity of the polymerase.
d. Mutations change other genes into DNA polymerases.

Back 30

c. The process of gene duplication would provide multiple copies of a gene with the function of synthesizing DNA. Over evolutionary time, variation among these genes may arise due to the accumulation of mutations.

Front 31

Point mutations that do not alter the amino acid sequence of the resulting gene product are called _____________ mutations.
a. frameshift
b. natural
c. silent
d. nonsense
e. missense

Back 31

c. Silent mutation do not alter the polypeptide chain.

Front 32

Some point mutations will lead to an mRNA that produces a much short polypeptide. This type of mutation is known as a __________ mutation.
a. neutral
b. silent
c. missense
d. nonsense
e. chromosomal

Back 32

d. Nonsense mutation change a normal codon to a stop codon, resulting in a shorter polypeptide chain.

Front 33

The type of mutation that alters the entire amino acid sequence from the site of the mutation is known as a __________ mutation.
a. neutral
b. silent
c. missense
d. nonsense
e. frameshift

Back 33

e. Frameshift mutation are caused by the insertion or deletion of nucleotides. This changes the reading frame of the gene and thus the entire amino acid sequence from the point of the mutation.

Front 34

Mutagens can cause mutations by
a. chemically altering DNA nucleotides.
b. disrupting DNA replication.
c. altering the genetic code of an organism.
d. all of the above.
e. a and b only.

Back 34

e. Mutagens can chemically or physically alter the DNA, thus changing the nucleotide sequence in existing or newly synthesized DNA molecules.

Front 35

The mutagenic effect of UV light is
a. the alteration of cytosine bases to adenine bases.
b. the formation of purine dimers that interfere with genetic expression.
c. the breaking of the sugar-phosphate backbone of the DNA molecule.
d. the formation of pyrimidine dimers that disrupt DNA replication.
e. the deletion of thymine bases along the DNA molecule.

Back 35

d. UV light causes the formation of pyrimidine dimers. During replication, DNA polymerase is likely to cause a mismatch when reading the area of the DNA strand with a thymine dimer.

Front 36

The Ames test
a. provides a way to determine if any type of cell has experienced a mutation.
b. provides a way to determine the mutagenic effect of certain types of agents.
c. allows researchers to experimentally disrupt gene activity by causing a mutation in a specific gene.
d. provides a way to repair mutations in bacterial cells.
e. all of the above.

Back 36

b. The Ames test determines the mutagenic effects of certain agents.

Front 37

Xeroderma pigmentosum
a. is a genetic disorder that results in uncontrolled cell growth.
b. is a genetic disorder where normal NER systems are not fully functional.
c. is a genetic disorder that results in the loss of pigment in certain patches of skin.
d. results from the lack of DNA polymerase proofreading.
e. both b and d.

Back 37

b. XP is the result of defects in NER system proteins.

Front 38

During mismatch repair, the parental strand is distinguishable from the new strand by
a. the lack of mutations in the parental strand.
b. the presence of methyl groups on the new strand.
c. the presence of methyl groups on the parental strand.
d. the 3’ to 5’ orientation of the strand.
e. the AUG codon on the new strand.

Back 38

c. The parental strand is detectable by the presence of methyl groups.

Front 39

Cancer cells are said to be metastatic when they
a. begin to divide uncontrollably.
b. invade healthy tissue.
c. migrate to other parts of the body.
d. cause mutations in other healthy cells.
e. all of the above.

Back 39

c. The term metastasis refers to the migration or movement of cancer cells to other parts of the body.

Front 40

Oncogenes are
a. mutations in genes that normally inhibit the progression of a cell through the cell cycle.
b. mutations that cause the overexpression of genes that normally stimulated cell division.
c. viruses that cause cancer.
d. mutations in genes that cause metastasis.
e. all of the above.

Back 40

b. Oncogenes are mutated forms of proto-oncogenes. Normally, proto-oncogenes stimulate cell division. However, mutations to oncogenes result in overactive expression of these genes and uncontrolled cell growth.

Front 41

Which of the following best represents the central dogma of gene expression?
a. During transcription, DNA codes for polypeptides.
b. During transcription, DNA codes for RNA, which codes for polypeptides during translation.
c. During translation, DNA codes for RNA, which codes for polypeptides during transcription.
d. None of the above.

Back 41

b. The proper sequence for gene expression in both prokaryotes and eukaryotes is transcription followed by translation. During transcription, a portion of the DNA molecule serves as a template for a complementary mRNA. The mRNA is then translated to produce a polypeptide.

Front 42

Transcription of a gene begins at a site on DNA called ________ and ends at a site on DNA known as ___________.
a. an initiation codon, termination codon
b. a promoter, the termination codon
c. an initiation codon, the terminator
d. a promoter, the terminator
e. an initiator, the terminator

Back 42

d. A promoter is a DNA sequence that signals the initiation site of transcription. A terminator is a DNA sequence that signals the end of the gene sequence. The term codon refers to RNA triplets, not DNA.

Front 43

The product of transcription of a structural gene is
a. tRNA.
b. mRNA.
c. rRNA.
d. polypeptide.
e. a, b and c.

Back 43

b. The term structural gene refers to a DNA sequence that codes for an amino acid sequence; therefore, the product of the transcription would be mRNA. rRNA and tRNA do not code for amino acid sequences and would not be transcribed from a structural gene.

Front 44

During eukaryotic RNA processing, the nontranslated sequences that are removed are called
a. exons.
b. introns.
c. promoters.
d. codons.
e. ribozymes.

Back 44

b. The term “introns” refers to intervening sequences. These are not part of the final mRNA and are removed.

Front 45

Ribozymes are
a. the organelles where translation takes place.
b. the RNA molecules that are components of ribosomes.
c. the proteins that are components of ribosomes.
d. the portions of the pre-mRNA that are removed.
e. RNA molecules that catalyze chemical reactions.

Back 45

e. Ribozymes are RNA molecules that have catalytic abilities. They function in splicing of some introns as well as the reaction that links the amino acids during translation.

Front 46

The _____________ is the organelle where the translation process takes place.
a. mitochondria
b. nucleus
c. ribosome
d. lysosome
e. ribozyme

Back 46

c. The ribosome is the “workbench” of protein synthesis.

Front 47

The region of the tRNA that is complementary to the triplet on the mRNA is
a. the acceptor stem.
b. the codon.
c. the peptidyl site.
d. the anticodon.
e. the adaptor loop.

Back 47

The region of the tRNA that is complementary to the triplet on the mRNA is
a. the acceptor stem.
b. the codon.
c. the peptidyl site.
d. the anticodon.
e. the adaptor loop.

Front 48

During the initiation step of translation, the first codon, ____ will enter the _________ and associate with the initiator tRNA.
a. UAG, A site
b. AUG, A site
c. UAG, P site
d. AUG, P site
e. AUG, E site

Back 48

d. AUG is the initiator, or start, codon. It associates with the initiator tRNA and aligns in the P site of the ribosome.

Front 49

The movement of the polypeptide from the tRNA in the P site to the rTNA in the A site is referred to as
a. peptide bonding.
b. aminoacyl binding.
c. translation.
d. peptidyl transfer reaction.
e. elongation.

Back 49

d. peptidyl transfer reaction

Front 50

The synthesis of a polypeptide occurs during which stage of translation?
a. initiation
b. elongation
c. termination
d. splicing

Back 50

b. elongation