Download test bank for molecular cell biology 7th edition by lodish

Test Bank for Molecular Cell Biology 7th Edition by Lodish
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Molecular Genetic Techniques

PART A: Linking Concepts and Facts
5.1

Genetic Analysis of Mutations to Identify and Study Genes

1. A mutation that changes a cysteine codon to a tryptophan codon is called a
a. missense mutation.
b. nonsense mutation.
c. frameshift mutation.
d. silent mutation.

Ans: a

2. Crossing of a homozygous wild type with a mutant that is heterozygous for a dominant mutation will
result in F1 progeny of which
a. all show the mutant phenotype.
b. half show the wild-type phenotype and half show the mutant phenotype.
c. three-fourths show the wild-type phenotype and one-fourth show the mutant phenotype. d. all show
the wild-type phenotype.


Ans: b

3. A mutation in one gene that counteracts the effects of a mutation in another gene is known as a a.
temperature-sensitive mutation.
b. recessive mutation.
c. conditional mutation.

d. suppressor mutation.

Ans: d

5.2

DNA Cloning and Characterization

4. Which of the following enzymes will produce a blunt end (the cut site is indicated by the * in the
recognition sequence)?
a. TaqI (T*CGA)
b. EagI (C*GGCCG)
c. EcoRV (GAT*ATC)
d. NsiI (ATGCA*T)

Ans: c

5. DNA ligase
a. synthesizes DNA from an RNA template.


b. forms a phosphodiester bond.
c. joins Okazaki fragments.
d. b and c

Ans: d

6. Which of the following is a functional element of a plasmid?
a. origin of replication
b. drug-resistance gene

c. polylinker sequence
d. all of the above

Ans: d

7. Next generation sequencing is much more efficient than the Sanger method because

a.

it uses an RNA template instead of DNA.

b.

it uses gel electrophoresis to resolve end-labeled strands of DNA.

c.

it uses PCR amplification.

d.

b and c

Ans. c


5.3

Using Cloned DNA Fragments to Study Gene Expression

8. A DNA sequencing reaction contains which of the following?
a. dideoxyribonucleoside triphosphates
b. deoxyribonucleoside triphosphates
c. DNA polymerase
d. all of the above

Ans: d

9. The polymerase chain reaction (PCR) technique can be used for
a. direct isolation of a specific segment of genomic DNA.
b. preparation of probes.
c. synthesis of RNA from genomic DNA.
d. a and b

Ans: d

10. Southern blotting is used to detect a specific
a. DNA.
b. RNA.
c. protein.


d. carbohydrate.

Ans: a

11. In the large-scale production of a particular human protein in E. coli cells, the cDNA corresponding to
the protein was modified so that the expressed protein would have six histidine residues at the
Cterminus. The purpose of this modification was to
a.

facilitate transfer of the cDNA into the E. coli cells.

b.
c.

provide a promoter for the transcription of the cDNA in E. coli.
facilitate purification of the expressed protein though binding to an affinity column containing
chelated nickel atoms.

d.

prevent degradation of the expressed protein by E. coli proteases.

Ans: c

12. In situ hybridization is a powerful tool used in gene expression studies because it provides the
investigator with information pertaining to

a.

the cellular and tissue-specific localization of the mRNA encoded by a particular gene.

b.

the activity of the protein translated from a particular mRNA.

c.

the size of the mRNA transcript.

d.

all of the above

5.4 Identifying and Locating Human Disease Genes


13. The disease gene for cystic fibrosis is
a. autosomal dominant.
b. sex-linked dominant.
c. autosomal recessive.
d. sex-linked recessive.

Ans: c

14. A mutation that changes the recognition sequence for the restriction enzyme EcoRI from GAATTC to
GATTTC is an example of a

a. single nucleotide polymorphism (SNP).
b. simple sequence repeat (SSR).
c. a and b
d. all of the above

Ans: a

15. The identification of disease genes can be made more complicated because of a.
genetic heterogeneity.
b. location of the gene on the X chromosome.
c. polygenic traits.

d. a and c


Ans: d

16. Linkage studies can map disease genes with a resolution of about one centimorgan. Typically, a DNA
region this size could contain about _______ genes.
a.

1 or 2

b.

10–50

c.

100–200

d.

1000–2000

Ans: b

17. A haplotype is a set of closely linked genetic markers on a particular chromosome that tend to be
inherited together. The genetic technique that looks at inheritance patterns and uses haplotypes in
determining gene locations is
a.
linkage mapping.

b.

linkage disequilibrium mapping.

c.

candidate gene approach.

d.

all of the above

Ans: b

5.5 Inactivating the Function of Specific Genes in Eukaryotes

18. A step in the generation of knockout mutations in mice includes selection of embryonic stem (ES)
cells that are


a. resistant to G418 and resistant to ganciclovir.
b. sensitive to G418 and resistant to ganciclovir.
c. resistant to G418 and sensitive to ganciclovir.
d. sensitive to G418 and sensitive to ganciclovir.

Ans: a

19. What method can be used to functionally inactivate a gene without altering its sequence? a.
gene knockout
b. RNA interference

c. dominant negative mutation
d. b and c

Ans: d

20. In RNA interference studies, the double-stranded RNA
a. disrupts the target DNA sequence.
b. results in the destruction of the target mRNA.
c. destroys the target protein.
d. all of the above

Ans: b


21. To study the function of the essential cytosolic Hsc70 genes in yeast, researchers constructed a
shuttle vector in which a copy of the Hsc70 gene was ligated to the GAL1 promoter. The vector was then
introduced into haploid yeast cells in which all four copies of the Hsc70 genes had been disrupted.
Following introduction of the vector, you would expect that
a.

the yeast cells would grow on both glucose and galactose media.

b.

the yeast cells would grow on glucose but not galactose medium.

c.

the yeast cells would grow on galactose but not glucose medium.

d.

on transfer to either glucose or galactose medium, the vector-carrying cells would eventually
stop growing because of insufficient Hsc70 activity.

Ans: c

22. To inactivate the function of a wild-type small GTPase one could introduce
a.

a dominant negative allele of a GTPase gene that binds to and inactivates a guanine nucleotide
exchange factor.

b.

an RNAi to silence the expression of the guanine nucleotide exchange factor.

c.

Cre recombinase.

d.

a and c

Ans. d

PART B: Testing on the Concepts

5.1

Genetic Analysis of Mutations to Identify and Study Genes


23.

Describe how genetic complementation can be used in yeast to determine whether two
different recessive mutations are in the same or different genes.

Ans: Yeasts are normally a haploid organism but can be made to form a diploid organism by mating.
Genetic complementation is a phenomenon whereby the wild-type phenotype can be restored after
mating two recessive mutants. If two recessive mutations are in the same gene then a diploid organism
containing both mutations will show a mutant phenotype because neither allele provides a functional
copy of the gene. If two recessive mutations are in different genes then the diploid yeast will show a
wild-type phenotype because a wild-type allele of each gene will be present.

24.

Describe the properties and utility of temperature-sensitive mutations.

Ans: A temperature-sensitive mutation is a mutation that expresses a wild-type phenotype at one
wild

-sensitive or
temperature and a mutant phenotype at another temperature. For example, the protein may exist in a
conditional mutations are especially useful for isolation of mutations in essential genes. In this case,
cells with the mutation can be propagated normally at the permissive temperature and the effect of the
mutation can be studied at the nonpermissive temperature.

5.2

DNA Cloning and Characterization

25.

Describe some typical features of a restriction enzyme recognition sequence.

Ans: Restriction enzyme recognition sequences are typically 4–8 base pairs in length and are
palindromic, which means that the sequence read in the 5´ to 3´ direction is the same on each DNA


strand. For example, the recognition sequence for EcoRI is 5´ GAATTC 3´. The complementary sequence
is 3´ CTTAAG 5´, which is the same sequence when read in the 5´ to 3´ direction.

26.

Describe the essential features of a yeast shuttle vector.

Ans: A yeast shuttle vector is a plasmid that can replicate in both bacteria (E. coli) and yeast. For
replication in bacteria, the plasmid needs a bacterial origin of replication and a gene (e.g., for resistance
to ampicillin) for selection in transformed E. coli. For replication in yeast, the plasmid needs a yeast
origin of replication (autonomously replicating sequence, ARS), a gene (e.g., ura3) for selection in
transformed yeast, and a centromere. In addition, a polylinker sequence for the efficient cloning of
foreign DNA is necessary.

5.3

Using Cloned DNA Fragments to Study Gene Expression

27.

What is epitope tagging? What is its application?

Ans: Epitope tagging is a method by which a cloned cDNA sequence is modified by the addition of a
short DNA sequence that encodes a peptide recognized by a known monoclonal antibody. The short
encoded peptide is called the epitope. This modified cDNA can be introduced and expressed in cells and
the location of the modified protein can be determined immunologically. Because the modified protein
now expresses the epitope, the monoclonal antibody to the epitope can be used to detect the presence
of the epitope-tagged protein. Epitope tagging allows the use of a single antibody for detection of any
epitope-modified protein and eliminates the need to generate an antibody to each protein of interest.

28.

Compare the advantages and limitations of microarrays and Northern blots for analyzing gene
expression.


Ans: Microarrays allow a more global analysis of gene expression by analyzing thousands of genes
simultaneously. Using the cluster-analysis program, groups of known and unknown genes that are
regulated in a coordinated fashion can be revealed. Northern blots allow the analysis of only a few
genes at a time; however, a Northern blot can reveal more information about the RNAs than a
microarray. For example, a Northern blot can reveal the presence of multiple mRNAs that may be
differentially expressed. The presence of multiple mRNAs could be missed by microarray analysis.

5.4

Identifying and Localizing Human Disease Genes

29.

How can linkage analysis position genes on a chromosome?

Ans: Linkage analysis examines the frequency of genetic recombination between two genes or markers
on a chromosome. A genetic map or linkage map contains many markers mapped to a chromosome.
Linkage analysis between the unknown gene and one of the known markers allows the rough
localization of the gene on the chromosome. The basis for recombinational analysis is that two genes
that are far apart on a chromosome will have a higher frequency of recombination than two genes that
are close together. Thus, if recombination between the gene of interest and a marker is very low, then
the gene is likely located near that marker gene.

30.

How does genetic heterogeneity or polygenic traits make the identification of a disease gene
more difficult?

globin chain of hemoglobin causes sickle-cell anemia. However, some diseases result from mutations in
not just one but multiple genes in an individual (a polygenic trait), whereas others result from mutations
in any one of multiple different genes in different individuals, a phenomenon known as genetic
heterogeneity. When more than one gene is involved in a single individual or among individuals, then
mapping of the disease gene to a marker gene on a linkage map becomes much more complex.

5.5 Inactivating the Function of Specific Genes in Eukaryotes


31.

RNAi is used to functionally inactivate genes in cells and whole organisms like C. elegans.
Describe the basics of how you would knock down the expression of a gene required for muscle
formation in C. elegans and what method could you use to confirm that your results were
specifically attributed to the RNAi?

Ans: The cDNA coding sequence of the muscle target gene is cloned, in both the sense and antisense
orientations, adjacent to a strong promoter into plasmids. In vitro transcription of both constructs with
RNA polymerase and rNTPs yield RNA copies complementary to the coding and antisense sequences.
These copies are allowed to anneal and then microinjected into the gonad of C. elegans. In the
developing embryos, Dicer cleaves the double-stranded RNA into siRNAs, which bind to the
corresponding endogenous mRNA target. Binding leads either to the degradation of the message or it
serves to prevent the translation of the mRNA into a functional protein. In situ hybridization, Northern
blot, or RT-PCR analyses are used to detect changes in the expression of the mRNA transcript, whereas
immunocytochemistry or Western blot analysis are used to detect changes at the protein level.

32.

How can the function of an essential gene required for embryonic development be studied in
an adult knockout mouse?

Ans: A standard knockout mouse cannot be created to study an essential embryonic gene because the
loss of both copies of the gene would lead to embryonic death. The loxP-Cre system, however, can be
employed to generate a conditional knockout mouse. In this case, the essential embryonic gene would
be flanked with loxP sites. A transgenic mouse homozygous for the loxP modified embryonic gene
would be mated with a mouse heterozygous for the loxP modified embryonic gene, which also
expresses the Cre protein. The Cre protein would be expressed from a development-specific promoter
that turns on Cre expression in a neonatal or adult mouse, leading to loss of the target gene only in
postembryonic mice.