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: 1022 yellow and 3001
green (4023 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: A
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
3) Karl is an albino man. He is married to Irene, a woman of normal pigmentation, and they have
two children. One of the children is of normal pigmentation and one child is albino (without
melanin pigmentation). Albinism is an autosomal (not sex-linked) recessive trait. What is the
genotype of Irene?
A) Irene is homozygous for the albino trait
B) Irene is homozygeous for the normal pigmentation.
C) Irene is heterozygeous.
Answer: C
4) A man will pass his X-linked genes 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
Answer: A
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 amino acids that make up the DNA molecule contain the information needed to make
cellular proteins.
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C) The base sequence of DNA carries all the information needed to code for mRNA and
eventually for proteins.
Answer: C
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
Figure 14.2
7) Identify the leading strand during duplication of DNA in Figure 14.2.
A) a
B) b
C) c
D) d
Answer: A
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
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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 arg3
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: C
10) In the process of translation, _____.
A) DNA is replicated
B) RNA is synthesized
C) proteins are synthesized
Answer: C
11) Under which conditions would it post-translational control be useful? And what would be the
downside of this?
A) Post-translational control doesn’t cost a lot of energy, and is useful when a high efficiency of
control is required. The downside is that it is relatively slow.
B) Post-translational control is very fast and is useful when a quick response to a changing
environment is required. The downside is that it requires a lot of energy.
Ans: B
12) Codons, the three base sequences that code for specific amino acids, are part of _____.
A) proteins
B) RNA polymerase
C) ribosomes
D) mRNA
Answer: D
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
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D) describes a series of catalytic reactions
Answer: A
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
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
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
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
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
19) Which method is utilized by eukaryotes to control their gene expression that is not used in
bacteria?
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A) control of RNA splicing
B) transcriptional control
C) post-translational control
D) translational control
Answer: A
20) Which of the following allows more than one type of protein to be produced from one gene?
A) alternative forms of chromatin remodeling
B) alternative forms of RNA splicing
C) control of the frequency of translation initiation
D) all of the above
Answer: B
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
BONUS 2) You make a karyotype of a the cells of a patient and find aneuploidy and
chromosome translocations. In other words, you find _______________, which that the patient
likely has _________________
A) Point mutations; cancer
B) Point mutations; lactose intolerance
C) Chromosome mutations; lactose intolerance
D) Chromosome mutations; cancer
E) Silent and neutral mutations; very high fitness
Ans: D
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