Download Biological Science, 4e (Freeman)

1) Mendel crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to
self-pollinate to produce an F2 generation. The results were as follows: 6022 yellow and 2001
green (8023 total). The allele for green seeds has what relationship to the allele for yellow seeds?
A) dominant
B) incomplete dominant
C) recessive
D) codominant
Answer: C
Reference: Section 13.1
Bloom's Taxonomy: Level 4 Analysis
2) A man and woman are both of normal pigmentation, but both have one parent who is albino
(without melanin pigmentation). Albinism is an autosomal (not sex-linked) recessive trait. What
is the probability that their first child will be an albino?
A) 0
B) 1/8
C) 1/2
D) 1/4
E) 1
Answer: D
Reference: Section 13.3
Bloom's Taxonomy: Level 3 Application
3) A man and woman are both of normal pigmentation and have a one child out of three who is
albino (without melanin pigmentation). Albinism is an autosomal (not sex-linked) recessive trait.
What are the genotypes of the albino's parents?
A) One parent must be homozygous for the recessive allele; the other parent can be homozygous
dominant, homozygous recessive, or heterozygous.
B) One parent must be heterozygous; the other parent can be homozygous dominant,
homozygous recessive, or heterozygous.
C) Both parents must be heterozygous.
D) One parent must be homozygous dominant; the other parent must be heterozygous.
E) Both parents must be homozygous dominant.
Answer: C
Reference: Section 13.3
Bloom's Taxonomy: Level 3 Application
4) A man who carries an allele of an X-linked gene will pass it on to _____.
A) all of his daughters
B) half of his daughters
C) all of his sons
D) half of his sons
E) all of his children
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Answer: A
Reference: Section 13.4
Bloom's Taxonomy: Level 2 Comprehension
5) How does the simple primary and secondary structure of DNA hold the information needed to
code for the many features of multicellular organisms?
A) The hydrogen bonding among backbone constituents carries coded information.
B) The base sequence of DNA carries all the information needed to code for proteins.
C) The width of the double helix changes at each gene due to differences in hydrogen bonds.
D) The amino acids that make up the DNA molecule contain the information needed to make
cellular proteins.
Answer: B
Reference: Section 14.2
Bloom's Taxonomy: Level 3 Application
6) What is the difference between the leading strand and the lagging strand in DNA replication?
A) The leading strand is synthesized in the 3′→5′ direction in a discontinuous fashion, while the
lagging strand is synthesized in the 5′→3′ direction in a continuous fashion.
B) The leading strand requires an RNA primer, whereas the lagging strand does not.
C) The leading strand is synthesized continuously in the 5′→3′ direction, while the lagging
strand is synthesized discontinuously in the 5′→3′ direction.
D) There are different DNA polymerases involved in elongation of the leading strand and the
lagging strand.
Answer: C
Reference: Sections 14.1,14.3
Bloom's Taxonomy: Level 3 Application
Figure 14.2
7) Identify the lagging strand during duplication of DNA in Figure 14.2.
A) a
B) b
C) c
D) d
Answer: C
Reference: Sections 14.1,14.3
Bloom's Taxonomy: Level 3 Application
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8) Which of the following are important in reducing the errors in DNA replication in E. coli
organisms?
A) proofreading activity of the epsilon subunit of DNA polymerase III
B) mismatch repair
C) nucleotide excision repair
D) All of the above minimize errors in DNA replication in E. coli.
Answer: D
Reference: Section 14.5
Bloom's Taxonomy: Level 3 Application
Mutant Strain Medium with:
No ornithine
No citrulline
No arginine
arg1
No growth
arg2
No growth
arg3
No growth
Medium with:
Ornithine
No citrulline
No arginine
GROWTH
No growth
No growth
Medium with:
No ornithine
Citrulline
No Arginine
GROWTH
GROWTH
No growth
Medium with:
No ornithine
No citrulline
Arginine
GROWTH
GROWTH
GROWTH
9) According to the table and figure above, which enzyme is defective in the strain with the arg2
mutation?
A) the enzyme that converts the precursor to ornithine
B) the enzyme that converts ornithine to citrulline
C) the enzyme that converts citrulline to arginine
D) the enzyme that converts the precursor to citrulline
Answer: B
Reference: Section 15.1
Bloom's Taxonomy: Level 3 Application
10) Knockout mice have been genetically altered to knock out specific genes. How are these
mice most often used in research?
A) to study DNA replication in the defective genes (those that have been altered)
B) to determine the role of proteins coded for by those genes that are knocked out
C) to examine defects in DNA structure in those regions that have been altered
D) to study the effect of radiation on DNA
Answer: B
Reference: Section 15.1
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11) In the process of transcription, _____.
A) DNA is replicated
B) RNA is synthesized
C) proteins are synthesized
D) mRNA attaches to ribosomes
Answer: B
Reference: Section 15.2
Bloom's Taxonomy: Level 2 Comprehension
12) Codons, the three base sequences that code for specific amino acids, are part of _____.
A) protein
B) mRNA
C) tRNA
D) rRNA
Answer: B
Reference: Section 15.3
Bloom's Taxonomy: Level 2 Comprehension
13) The statement, DNA → RNA → Proteins, _____.
A) has become known as the central dogma
B) depicts the regulation of gene expression
C) never varies in any organism, virus, or prion
D) describes a series of catalytic reactions
Answer: A
Reference: Section 15.2
Bloom's Taxonomy: Level 2 Comprehension
14) Which of the following findings violated the central dogma?
A) the discovery of RNA viruses that synthesize DNA using reverse transcriptase
B) the discovery that the Archaea and Bacteria are more distantly related than are Archaea and
Eukarya
C) the discovery of ribozymes
D) the discovery of DNA as the unit of genetic inheritance
Answer: A
Reference: Section 15.2
Bloom's Taxonomy: Level 3 Application
15) Ribosomes can attach to prokaryotic (like bacteria!) messenger RNA _____.
A) once post-transcriptional modification is complete
B) before transcription is complete
C) once replication is complete
D) once the primary transcript has been released from RNA polymerase
Answer: B
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Reference: Section 16.3
Bloom's Taxonomy: Level 2 Comprehension
16) How does termination of translation take place?
A) The end of the mRNA molecule is reached.
B) A stop codon is reached.
C) The 5' cap is reached.
D) The poly A tail is reached.
Answer: B
Reference: Section 16.5
Bloom's Taxonomy: Level 3 Application
17) Extracellular glucose inhibits transcription of the lac operon by _____.
A) strengthening the binding of repressor to the operator
B) weakening the binding of repressor to the operator
C) inhibiting RNA polymerase from opening the strands of DNA to initiate transcription
D) reducing the levels of intracellular cAMP
Answer: D
Reference: Section 17.4
Bloom's Taxonomy: Level 3 Application
18) An E. coli cell without a functional lacI gene is expected to _____.
A) never produce β-galactosidase
B) always produce β-galactosidase
C) be unable to transport lactose into the cell
D) be unable to metabolize lactose within the cell
Answer: B
Reference: Section 17.2
Bloom's Taxonomy: Level 3 Application
19) Which method(s) is/are utilized by eukaryotes to control their gene expression that is/are 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
E) control of chromatin remodeling, RNA splicing, and transcription
Answer: D
Reference: Section 18.1
Bloom's Taxonomy: Level 2 Comprehension
20) Which of the following allows more than one type of protein to be produced from one gene?
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A) alternative forms of chromatin remodeling
B) alternative forms of RNA splicing
C) alternative forms of nucleosomes
D) control of the frequency of translation initiation
E) all of the above
Answer: B
Reference: Section 18.4
Bloom's Taxonomy: Level 1 Knowledge
BONUS 1) In humans, blue eyes are inherited as a recessive autosomal trait and color blindness
is an X-linked recessive trait. A woman with blue eyes and normal color vision whose father was
color blind marries a man who also has normal color vision. He has brown eyes, but his mother
had blue eyes. Which of the following do you expect to be true for their sons?
A) One-half of their sons will have normal color vision and brown eyes; 1/2 of their sons will
have normal color vision and blue eyes.
B) Their sons will all have normal color vision and brown eyes.
C) One-fourth of their sons will be color blind and have blue eyes, 1/4 of their sons will be color
blind and have brown eyes, 1/4 of their sons will have normal color vision and blue eyes, 1/4 of
their sons will have normal color vision and brown eyes.
D) Their sons will all have normal color vision and blue eyes.
E) One-half of their sons will be color blind and have blue eyes; 1/2 their sons will be color blind
and have brown eyes.
Answer: C
Reference: Section 13.6
Bloom's Taxonomy: Level 4 Analysis
BONUS 2) If a scientist wanted to prevent a regulatory protein from changing gene expression,
he must prevent physical contact between the protein and _____.
A) ribosomes
B) DNA
C) ribozymes
D) mRNA
Answer: B
Reference: Section 17.3
Bloom's Taxonomy: Level 4 Analysis
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