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WorksheetsAP Bio: Information Processing
Total questions: 78
Worksheet time: 2hrs 36mins
Name
Class
Date
1.
The product of the p53 gene ____.
a)
inhibits the cell cycle
b)
slows down the rate of DNA replication by interfering with the binding of DNA polymerase
c)
causes cells to reduce expression of genes involved in DNA repair
d)
allows cells to pass on mutations due to DNA damage
2.
At which phase are centrioles beginning to move apart in animal cells?
a)
anaphase
b)
prometaphase
c)
metaphase
d)
prophase
3.
If there are 20 centromeres in a cell at anaphase, how many chromosomes are there in each daughter cell following cytokinesis?
a)
10
b)
20
c)
40
d)
80
4.
Measurements of the amount of DNA per nucleus were taken on a large number of cells from a growing fungus. The measured DNA levels ranged from 3 to 6 picograms per nucleus. In which stage of the cell cycle did the nucleus contain 6 picograms of DNA?
a)
G1
b)
S
c)
G2
d)
M
5.
The cyclin component of MPF is destroyed toward the end of which phase?
a)
G1
b)
S
c)
G2
d)
M
6.
During which phase of mitosis do the chromatids become chromosomes?
a)
telophase
b)
anaphase
c)
prophase
d)
metaphase
7.
A cleavage furrow is ____.
a)
a ring of vesicles forming a cell plate
b)
the separation of divided prokaryotes
c)
a groove in the plasma membrane between daughter nuclei
d)
the space that is created between two chromatids during anaphase
8.
Through a microscope, you can see a cell plate beginning to develop across the middle of a cell and nuclei forming on either side of the cell plate. This cell is most likely ____.
a)
an animal cell in the process of cytokinesis
b)
a plant cell in the process of cytokinesis
c)
an animal cell in the S phase of the cell cycle
d)
a plant cell in metaphase
9.
MPF is a dimer consisting of ____.
a)
a growth factor and mitotic factor
b)
ATP synthetase and a protease
c)
cyclin and tubulin
d)
cyclin and a cyclin-dependent kinase
10.
What happens if MPF (mitosis-promoting factor) is introduced into immature frog oocytes that are arrested in G2?
a)
Nothing happens.
b)
The cells undergo meiosis.
c)
The cells enter mitosis.
d)
Cell differentiation is triggered.
11.
The M-phase checkpoint ensures that all chromosomes are attached to the mitotic spindle. If this does not happen, cells would most likely be arrested in ____.
a)
Telophase
b)
prophase
c)
prometaphase
d)
metaphase
12.
Which of the following is released by platelets in the vicinity of an injury?
a)
PDGF
b)
MPF
c)
cyclin
d)
Cdk
13.
Which of the following triggers the cell's passage past the G2 checkpoint into mitosis?
a)
PDGF
b)
MPF
c)
cyclin
d)
Cdk
14.
A research team began a study of a cultured cell line. Their preliminary observations showed them that the cell line did not exhibit either density-dependent inhibition or anchorage dependence. What could they conclude right away?
a)
The cells are unable to form spindle microtubules.
b)
They have altered the series of cell cycle phases.
c)
The cells show characteristics of tumors.
d)
They were originally derived from an elderly organism.
15.
Density-dependent inhibition is explained by which of the following?
a)
As cells become more numerous, they begin to squeeze against each other, restricting their size and ability to produce control factors.
b)
As cells become more numerous, the cell surface proteins of one cell contact the adjoining cells and they stop dividing.
c)
As cells become more numerous, the protein kinases they produce begin to compete with each other, such that the proteins produced by one cell essentially cancel those produced by its neighbor.
d)
As cells become more numerous, the level of waste products increases, eventually slowing down metabolism.
16.
Cells from advanced malignant tumors often have very abnormal chromosomes and an abnormal number of chromosomes. What might explain the association between malignant tumors and chromosomal abnormalities?
a)
Cancer cells are no longer density-dependent.
b)
Cancer cells are no longer anchorage-dependent.
c)
Cell cycle checkpoints are not in place to stop cells with chromosome abnormalities.
d)
Transformation introduces new chromosomes into cells.
17.
Exposure of zebrafish nuclei to meiotic cytosol resulted in phosphorylation of NEP55 and L68 proteins by cyclin-dependent kinase 2. NEP55 is a protein of the inner nuclear membrane, and L68 is a protein of the nuclear lamina. What is the most likely role of phosphorylation of these proteins in the process of mitosis?
a)
They enable the attachment of the spindle microtubules to kinetochore regions of the centromere.
b)
They are involved in the disassembly and dispersal of the nucleolus.
c)
They are involved in the disassembly of the nuclear envelope.
d)
They assist in the movement of the centrosomes to opposite sides of the nucleus.
18.
Sister chromatids separate from each other during ____.
a)
meiosis I only
b)
meiosis II only
c)
mitosis and meiosis I
d)
mitosis and meiosis II
19.
Somatic cells of roundworms have four individual chromosomes per cell. How many chromosomes would you expect to find in an ovum from a roundworm?
a)
four
b)
two
c)
eight
d)
a diploid number
20.
What is a major difference between mitosis and meiosis I in a diploid organism?
a)
Sister chromatids separate in mitosis, while homologous pairs of chromosomes separate in meiosis I.
b)
Sister chromatids separate in mitosis, while homologous pairs of chromosomes separate in meiosis II.
c)
DNA replication takes place prior to mitosis, but not before meiosis I.
d)
Only meiosis I results in daughter cells that contain identical genetic information.
21.
Crossing over normally takes place during which of the following processes?
a)
meiosis II
b)
meiosis I
c)
mitosis
d)
mitosis and meiosis II
22.
Centromeres split and sister chromatids migrate to opposite poles in meiosis ____.
a)
anaphase I
b)
telophase I
c)
anaphase II
d)
telophase II
23.
For a species with a haploid number of 23 chromosomes, how many different combinations of maternal and paternal chromosomes are possible for the gametes?
a)
23
b)
46
c)
about 1000
d)
about 8 million
24.
When homologous chromosomes cross over, what occurs?
a)
Two chromatids get tangled, resulting in one re-sequencing its DNA.
b)
Two sister chromatids exchange identical pieces of DNA.
c)
Corresponding segments of non-sister chromatids are exchanged.
d)
Maternal alleles are "corrected" to be like paternal alleles and vice versa.
25.
Mendel's second law of independent assortment has its basis in which of the following events of meiosis I?
a)
synapsis of homologous chromosomes
b)
crossing over
c)
alignment of tetrads at the equator
d)
separation of cells at telophase
26.
In the cross AaBbCc ´ AaBbCc, what is the probability of producing the genotype AABBCC?
a)
1/4
b)
1/8
c)
1/16
d)
1/64
27.
A man has extra digits (six fingers on each hand and six toes on each foot). His wife and their daughter have a normal number of digits. Having extra digits is a dominant trait. The couple's second child has extra digits. What is the probability that their next (third) child will have extra digits?
a)
1/2
b)
1/16
c)
1/8
d)
3/4
28.
Suppose two AaBbCc individuals are mated. Assuming that the genes are not linked, what fraction of the offspring are expected to be homozygous recessive for the three traits?
a)
1/4
b)
1/8
c)
1/16
d)
1/64
29.
Which of the following is an example of polygenic inheritance?
a)
pink flowers in snapdragons
b)
the ABO blood group in humans
c)
white and purple flower color in peas
d)
skin pigmentation in humans
30.
Which of the following provides an example of epistasis?
a)
Recessive genotypes for each of two genes (aabb) results in an albino corn snake.
b)
In rabbits and many other mammals, one genotype (ee) prevents any fur color from developing.
c)
In Drosophila (fruit flies), white eyes can be due to an X-linked gene or to a combination of other genes.
d)
In cacti, there are several genes for the type of spines.
31.
Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The flower color trait in radishes is an example of which of the following?
a)
a multiple allelic system
b)
sex linkage
c)
codominance
d)
incomplete dominance
32.
Feather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. Two blue budgies were crossed. Over the years, they produced twenty-two offspring, five of which were white. What are the most likely genotypes for the two blue budgies?
a)
yyBB and yyBB
b)
yyBB and yyBb
c)
yyBb and yyBb
d)
yyBb and yybb
33.
A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative. Which of the following is a possible phenotype for the father?
a)
A negative
b)
O negative
c)
AB negative
d)
B positive
34.
In birds, sex is determined by a ZW chromosome scheme. Males are ZZ and females are ZW. A recessive lethal allele that causes death of the embryo is sometimes present on the Z chromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male that is heterozygous for the lethal allele and a normal female?
a)
2:1 male to female
b)
1:2 male to female
c)
1:1 male to female
d)
3:1 male to female
35.
Sex determination in mammals is due to the SRY gene. Which of the following could allow a person with an XX karyotype to develop a male phenotype?
a)
the loss of the SRY gene from an autosome
b)
translocation of SRY to a X chromosome
c)
a person with an extra autosomal chromosome
d)
a person with one normal and one shortened (deleted) X
36.
n humans, clear gender differentiation occurs, not at fertilization, but after the second month of gestation. What is the first event of this differentiation?
a)
formation of testosterone in male embryos
b)
formation of estrogens in female embryos
c)
activation of SRY in male embryos and masculinization of the gonads
d)
activation of SRY in females and feminization of the gonads
37.
Pseudohypertrophic muscular dystrophy is a human disorder that causes gradual deterioration of the muscles. Only boys are affected, and they are always born to phenotypically normal parents. Due to the severity of the disease, the boys die in their teens. Is this disorder likely to be caused by a dominant or recessive allele? Is its inheritance sex-linked or autosomal?
a)
dominant, sex-linked
b)
recessive, autosomal
c)
recessive, sex-linked
d)
incomplete dominant, sex-linked
38.
What is an adaptive advantage of recombination between linked genes?
a)
Recombination is required for independent assortment.
b)
Recombination must occur or genes will not assort independently.
c)
New allele combinations are acted upon by natural selection.
d)
The forces on the cell during meiosis II results in recombination.
39.
If recombination frequency is equal to distance in map units, what is the approximate distance between genes A and B?
a)
3 map units
b)
6 map units
c)
15 map units
d)
30 map units
40.
The greatest distance among the three genes is between a and c. What does this mean?
a)
Gene c is between a and b.
b)
Genes are in the order: a—b—c.
c)
Gene a is not recombining with c.
d)
Gene a is between b and c.
41.
In Drosophila melanogaster, vestigial wings are caused by a recessive allele of a gene that is linked to a gene with a recessive allele that causes black body color. Morgan crossed black-bodied, normal-winged females and gray-bodied, vestigial-winged males. The F1 were all gray bodied, normal winged. The F1 females were crossed to homozygous recessive males to produce testcross progeny. Morgan calculated the map distance to be 17 map units. Which of the following is correct about the testcross progeny?
a)
black-bodied, normal-winged flies = 17% of the total
b)
black-bodied, normal-winged flies PLUS gray-bodied, vestigial-winged flies = 17% of the total
c)
gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total
d)
black-bodied, vestigial-winged flies = 17% of the total
42.
Which of the following is generally true of aneuploidies in newborns?
a)
A monosomy is more frequent than a trisomy.
b)
Monosomy X is the only viable monosomy known to occur in humans.
c)
Human aneuploidy usually conveys an adaptive advantage in humans.
d)
An aneuploidy resulting in the deletion of a chromosome segment is less serious than a duplication.
43.
Mitochondrial DNA is primarily involved in coding for proteins needed for protein complexes of the electron transport chain and ATP synthase. Therefore, mutations in mitochondrial genes would most affect ____.
a)
DNA synthesis in cells of the immune system
b)
the movement of oxygen into erythrocytes
c)
generation of ATP in muscle cells
d)
the storage of urine in the urinary bladder
44.
In his transformation experiments, what did Griffith observe?
a)
Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.
b)
Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic.
c)
Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.
d)
Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.
45.
For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work?
a)
There is no radioactive isotope of nitrogen.
b)
Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.
c)
Although there are more nitrogens in a nucleotide, labeled phosphates actually have sixteen extra neutrons; therefore, they are more radioactive.
d)
Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.
46.
In the polymerization of DNA, a phosphodiester bond is formed between a phosphate group of the nucleotide being added and ____ of the last nucleotide in the polymer.
a)
the 5' phosphate
b)
C6
c)
the 3' OH
d)
a nitrogen from the nitrogen-containing base
47.
In E. coli, there is a mutation in a gene called dnaB that alters the helicase that normally acts at the origin. Which of the following would you expect as a result of this mutation?
a)
Additional proofreading will occur.
b)
No replication fork will be formed.
c)
Replication will occur via RNA polymerase alone.
d)
Replication will require a DNA template from another source.
48.
How does the enzyme telomerase meet the challenge of replicating the ends of linear chromosomes?
a)
It adds a single 5' cap structure that resists degradation by nucleases.
b)
It causes specific double-strand DNA breaks that result in blunt ends on both strands.
c)
It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
d)
It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity.
49.
What is the function of topoisomerase?
a)
relieving strain in the DNA ahead of the replication fork
b)
elongating new DNA at a replication fork by adding nucleotides to the existing chain
c)
unwinding of the double helix
d)
stabilizing single-stranded DNA at the replication fork
50.
What provides the energy for the polymerization reactions in DNA synthesis?
a)
ATP
b)
DNA polymerase
c)
breaking the hydrogen bonds between complementary DNA strands
d)
the deoxyribonucleotide triphosphate substrates
51.
Telomere shortening puts a limit on the number of times a cell can divide. Research has shown that telomerase can extend the life span of cultured human cells. How might adding telomerase affect cellular aging?
a)
Telomerase will speed up the rate of cell proliferation.
b)
Telomerase eliminates telomere shortening and retards aging.
c)
Telomerase shortens telomeres, which delays cellular aging.
d)
Telomerase would have no effect on cellular aging.
52.
Researchers found E. coli that had mutation rates one hundred times higher than normal. Which of the following is the most likely cause of these results?
a)
The single-stranded binding proteins were malfunctioning.
b)
There were one or more mismatches in the RNA primer.
c)
The proofreading mechanism of DNA polymerase was not working properly.
d)
The DNA polymerase was unable to add bases to the end of the growing nucleic acid chain.
53.
If a cell were unable to produce histone proteins, which of the following would be a likely effect?
a)
There would be an increase in the amount of "satellite" DNA produced during centrifugation.
b)
The cell's DNA couldn't be packed into its nucleus.
c)
Spindle fibers would not form during prophase.
d)
Amplification of other genes would compensate for the lack of histones.
54.
The genetic code is essentially the same for all organisms. From this, one can logically assume which of the following?
a)
A gene from an organism can theoretically be expressed by any other organism.
b)
DNA was the first genetic material.
c)
The same codons in different organisms translate into different amino acids.
d)
Different organisms have different types of amino acids.
55.
"A produces B using enzyme A, then B produces C using enzyme B". is a simple metabolic pathway. According to Beadle and Tatum's hypothesis, how many genes are necessary for this pathway?
a)
1
b)
2
c)
3
d)
It cannot be determined from this pathway.
56.
Transcription in eukaryotes requires which of the following in addition to RNA polymerase?
a)
start and stop codons
b)
ribosomes and tRNA
c)
several transcription factors
d)
aminoacyl-tRNA synthetase
57.
Alternative RNA splicing ____.
a)
is a mechanism for increasing the rate of translation
b)
can allow the production of proteins of different sizes and functions from a single mRNA
c)
can allow the production of similar proteins from different RNAs
d)
increases the rate of transcription
58.
In the structural organization of many eukaryotic genes, individual exons may be related to which of the following?
a)
the sequence of the intron that immediately precedes each exon
b)
the number of polypeptides making up the functional protein
c)
the various domains of the polypeptide product
d)
the number of start sites for transcription
59.
Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the ____.
a)
binding of ribosomes to mRNA
b)
binding of the anticodon to small subunit of the ribosome
c)
attachment of amino acids to rRNAs
d)
binding of the anticodon to the codon and the attachment of amino acids to tRNAs
60.
What type of bonding is responsible for maintaining the shape of the tRNA molecule?
a)
ionic bonding between phosphates
b)
hydrogen bonding between base pairs
c)
van der Waals interactions between hydrogen atoms
d)
peptide bonding between amino acids
61.
The most commonly occurring mutation in people with cystic fibrosis is a deletion of a single codon. This results in ____.
a)
a base-pair substitution
b)
a frameshift mutation
c)
a polypeptide missing an amino acid
d)
a nonsense mutation
62.
If a scientist moves the promoter for the lac operon to the region between the beta galactosidase (lacZ) gene and the permease (lacY) gene, which of the following would be likely?
a)
The three structural genes will be expressed normally.
b)
RNA polymerase will no longer transcribe permease.
c)
The operon will still transcribe the lacZ and lacY genes, but the mRNA will not be translated.
d)
Beta galactosidase will not be produced.
63.
If a scientist moves the repressor gene (lacI), along with its promoter, to a position at some several thousand base pairs away from its normal position, we would expect the ____.
a)
repressor will no longer bind to the operator
b)
repressor will no longer bind to the inducer
c)
lac operon will be expressed continuously
d)
lac operon will function normally
64.
According to the lac operon model proposed by Jacob and Monod, what is predicted to occur if the operator is removed from the operon?
a)
The lac operon would be transcribed continuously.
b)
Only lacZ would be transcribed.
c)
Only lacY would be transcribed.
d)
Galactosidase permease would be produced, but would be incapable of transporting lactose.
65.
Which method is utilized by eukaryotes to control their gene expression that is NOT used in bacteria?
a)
control of chromatin remodeling
b)
control of RNA splicing
c)
transcriptional control
d)
control of both RNA splicing and chromatin remodeling
66.
The protein of the bicoid gene in Drosophila determines the ____ of the embryo.
a)
anterior-posterior axis
b)
anterior-lateral axis
c)
posterior-dorsal axis
d)
posterior-ventral axis
67.
Viruses use the host's machinery to make copies of themselves. However, some human viruses require a type of replication that humans do not normally have. For example, humans normally do not have the ability to convert RNA into DNA. How can these types of viruses infect humans, when human cells cannot perform a particular role that the virus requires?
a)
The virus causes mutations in the human cells, resulting in the formation of new enzymes that are capable of performing these roles.
b)
The viral genome codes for specialized enzymes not in the host.
c)
The virus infects only those cells and species that can perform all the replication roles necessary.
d)
Viruses can stay in a quiescent state until the host cell evolves this ability.
68.
The first class of drugs developed to treat AIDS, such as AZT, were known as reverse transcriptase inhibitors. They worked because they ____.
a)
targeted and destroyed the viral genome before it could be reverse transcribed into DNA
b)
bonded to the dsDNA genome of the virus in such a way that it could not separate for replication to occur
c)
bonded to the viral reverse transcriptase enzyme, thus preventing the virus from making a DNA copy of its RNA genome
d)
prevented host cells from producing the enzymes used by the virus to replicate its genome
69.
Which of the following human diseases is caused by a virus that requires reverse transcriptase to transcribe its genome inside the host cell?
a)
herpes
b)
AIDS
c)
smallpox
d)
influenza
70.
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 can incorporate a variety of nonstandard bases.
d)
RNA viruses are more sensitive to mutagens.
71.
Which of the following represents a difference between viruses and viroids?
a)
Viruses infect many types of cells, whereas viroids infect only prokaryotic cells.
b)
Viruses have capsids composed of protein, whereas viroids have no capsids.
c)
Viruses have genomes composed of RNA, whereas viroids have genomes composed of DNA.
d)
Viruses cannot pass through plasmodesmata, whereas viroids can.
72.
A person is most likely to recover from a viral infection if the infected cells ____.
a)
can undergo normal cell division
b)
can carry on translation, at least for a few hours
c)
produce and release viral protein
d)
transcribe viral mRNA
73.
Will treating a viral infection with antibiotics affect the course of the infection?
a)
No; antibiotics work by inhibiting enzymes specific to bacteria. Antibiotics have no effect on eukaryotic or virally encoded enzymes.
b)
No; antibiotics do not kill viruses because viruses do not have DNA or RNA.
c)
Yes; antibiotics activate the immune system, and this decreases the severity of the infection.
d)
Yes; antibiotics can prevent viral entry into the cell by binding to host-receptor proteins.
74.
Pax-6 is a gene that is involved in eye formation in many invertebrates, such as Drosophila. Pax-6 is also found in vertebrates. A Pax-6 gene from a mouse can be expressed in a fly and the protein (PAX-6) leads to a compound fly eye. This information suggests which of the following?
a)
Pax-6 genes are identical in nucleotide sequence.
b)
PAX-6 proteins have identical amino acid sequences.
c)
Pax-6 is highly conserved and shows shared evolutionary ancestry.
d)
PAX-6 proteins are different for formation of different kinds of eyes.
75.
Which of the following sequences is most likely to be cut by a restriction enzyme?
a)
AATTCT
TTAAGA
TTAAGA
b)
AATATT
TTATAA
TTATAA
c)
AAAATT
TTTTAA
TTTTAA
d)
ACTACT
TGATGA
TGATGA
76.
Which of the following is required to make complementary DNA (cDNA) from RNA?
a)
restriction enzymes (endonucleases)
b)
gene cloning
c)
DNA ligase
d)
reverse transcriptase
77.
Reproductive cloning of human embryos is generally considered unethical. However, on the subject of therapeutic cloning there is a wider divergence of opinion. Which of the following is a likely explanation?
a)
The use of adult stem cells is likely to produce more cell types than the use of embryonic stem cells.
b)
Cloning to produce embryonic stem cells may lead to great medical benefits for many.
c)
Cloning to produce stem cells relies on a different initial procedure than reproductive cloning.
d)
A clone that lives until the blastocyst stage does not yet have human DNA.
78.
Let us suppose that someone is successful at producing induced pluripotent stem cells (iPS) for replacement of pancreatic insulin-producing cells for people with type 1 diabetes. Which of the following could still be problems?
I. the possibility that, once introduced into the patient, the iPS cells produce nonpancreatic cells
II. the failure of the iPS cells to take up residence in the pancreas
III. the inability of the iPS cells to respond to appropriate regulatory signals
I. the possibility that, once introduced into the patient, the iPS cells produce nonpancreatic cells
II. the failure of the iPS cells to take up residence in the pancreas
III. the inability of the iPS cells to respond to appropriate regulatory signals
a)
I only
b)
II only
c)
III only
d)
I, II, and III
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