Unit 5 Practice Test
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) 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
A) Radioactive nitrogen has a half-life of 100,000 years, and the material would be too
dangerous for too long.
B) Avery et al. have already concluded that this experiment showed inconclusive results.
C) Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra
neutrons; therefore, they are more radioactive.
D) There is no radioactive isotope of nitrogen.
E) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not
distinguish between DNA and proteins.
2) Chargaff's analysis of the relative base composition of DNA was significant because he was able to
A) the human genome is more complex than that of other species.
B) the amount of ribose is always equivalent to deoxyribose.
C) the relative proportion of each of the four bases differs within individuals of a species.
D) transformation causes protein to be brought into the cell.
E) the amount of A is always equivalent to T, and C to G.
3) Mendel and Morgan did not know about the structure of DNA; however, which of the following of
their contributions was (were) necessary to Watson and Crick?
B) genetic distance and mapping
C) dominance vs. recessiveness
D) the particulate nature of the hereditary material
E) the usefulness of peas and Drosophila
4) What determines the nucleotide sequence of the newly synthesized strand during DNA
A) the primase used in the reaction
B) the particular DNA polymerase catalyzing the reaction
C) the arrangement of histones in the sugar phosphate backbone
D) the relative amounts of the four nucleoside triphosphates in the cell
E) the nucleotide sequence of the template strand
5) Which of the following statements describes the eukaryotic chromosome?
A) The nucleosome is its most basic functional subunit.
B) Active transcription occurs on heterochromatin.
C) It consists of a single linear molecule of double-stranded DNA.
D) The number of genes on each chromosome is different in different cell types of an organism.
E) It is composed of DNA alone.
6) The nitrogenous base adenine is found in all members of which group?
A) proteins, ATP, and DNA
B) ATP, RNA, and DNA
C) proteins, triglycerides, and testosterone
D) alpha glucose, ATP, and DNA
E) proteins, carbohydrates, and ATP
7) Using RNA as a template for protein synthesis instead of translating proteins directly from the
DNA is advantageous for the cell because
A) tRNA, rRNA and others are not transcribed.
B) RNA is much more stable than DNA.
C) RNA acts as an expendable copy of the genetic material.
D) only one mRNA molecule can be transcribed from a single gene, lowering the potential rate of
E) mRNA molecules are subject to mutation but DNA is not.
The following questions refer to Figure 17.2, a table of codons.
8) A peptide has the sequence NH2-phe-pro-lys-gly-phe-pro-COOH. Which of the following
sequences in the coding strand of the DNA could code for this peptide?
A) 5' ACT-TAC-CAT-AAA-CAT-TAC-UGA
B) 3' AUG-AAA-GGG-TTT-CCC-AAA-GGG
C) 5' TTT-CCC-AAA-GGG-TTT-CCC
D) 5' GGG-AAA-TTT-AAA-CCC-ACT-GGG
E) 3' UUU-CCC-AAA-GGG-UUU-CCC
9) What are the coding segments of a stretch of eukaryotic DNA called?
10) Choose the answer that has these events of protein synthesis in the proper sequence.
1. An aminoacyl-tRNA binds to the A site.
2. A peptide bond forms between the new amino acid and a polypeptide chain.
3. tRNA leaves the P site, and the P site remains vacant.
4. A small ribosomal subunit binds with mRNA.
5. tRNA translocates to the P site.
11) The role of a metabolite that controls a repressible operon is to
A) bind to the repressor protein and inactivate it.
B) bind to the repressor protein and activate it.
C) bind to the promoter region and decrease the affinity of RNA polymerase for the promoter.
D) bind to the operator region and block the attachment of RNA polymerase to the promoter.
E) increase the production of inactive repressor proteins.
12) When this is taken up by the cell, it binds to the repressor so that the repressor no longer binds to
13) Genomic imprinting, DNA methylation, and histone acetylation are all examples of
B) epigenetic phenomena.
C) chromosomal rearrangements.
D) genetic mutation.
14) Gene expression might be altered at the level of post-transcriptional processing in eukaryotes
rather than prokaryotes because of which of the following?
A) Prokaryotic genes are expressed as mRNA, which is more stable in the cell.
B) Eukaryotic coded polypeptides often require cleaving of signal sequences before localization.
C) Eukaryotic mRNAs get 5' caps and 3' tails.
D) Eukaryotic exons may be spliced in alternative patterns.
E) Prokaryotes use ribosomes of different structure and size.
15) In humans, the embryonic and fetal forms of hemoglobin have a higher affinity for oxygen than
that of adults. This is due to
A) identical genes that generate many copies of the ribosomes needed for fetal globin
B) pseudogenes, which interfere with gene expression in adults.
C) histone proteins changing shape during embryonic development.
D) the attachment of methyl groups to cytosine following birth, which changes the type of
E) nonidentical genes that produce different versions of globins during development.
Use the following information to answer the questions below.
A eukaryotic gene has "sticky ends" produced by the restriction endonuclease EcoRI. The gene is added to a mixture
containing EcoRI and a bacterial plasmid that carries two genes conferring resistance to ampicillin and tetracycline. The
plasmid has one recognition site for EcoRI located in the tetracycline resistance gene. This mixture is incubated for several
hours, exposed to DNA ligase, and then added to bacteria growing in nutrient broth. The bacteria are allowed to grow
overnight and are streaked on a plate using a technique that produces isolated colonies that are clones of the original.
Samples of these colonies are then grown in four different media: nutrient broth plus ampicillin, nutrient broth plus
tetracycline, nutrient broth plus ampicillin and tetracycline, and nutrient broth without antibiotics.
16) A principal problem with inserting an unmodified mammalian gene into a bacterial plasmid, and
then getting that gene expressed in bacteria, is that
A) prokaryotes use a different genetic code from that of eukaryotes.
B) bacterial RNA polymerase cannot make RNA complementary to mammalian DNA.
C) bacteria cannot remove eukaryotic introns.
D) bacteria translate polycistronic messages only.
E) bacterial DNA is not found in a membrane-bounded nucleus and is therefore incompatible
with mammalian DNA.
17) A gene that contains introns can be made shorter (but remain functional) for genetic engineering
purposes by using
A) a restriction enzyme to cut the gene into shorter pieces.
B) DNA ligase to put together fragments of the DNA that codes for a particular polypeptide.
C) DNA polymerase to reconstruct the gene from its polypeptide product.
D) reverse transcriptase to reconstruct the gene from its mRNA.
E) RNA polymerase to transcribe the gene.
18) Why are yeast cells frequently used as hosts for cloning?
A) they easily form colonies
B) they do not have plasmids.
C) only yeast cells allow the gene to be cloned
D) they can remove exons from mRNA.
E) they are eukaryotic cells
19) The segment of DNA shown in Figure 20.2 has restriction sites I and II, which create restriction
fragments A, B, and C. Which of the gels produced by electrophoresis shown below best represents
the separation and identity of these fragments?
20) Which of the following tools of recombinant DNA technology is incorrectly paired with its use?
A) reverse transcriptase–production of cDNA from mRNA
B) electrophoresis–separation of DNA fragments
C) DNA ligase–enzyme that cuts DNA, creating the sticky ends of restriction fragments
D) restriction enzyme–production of RFLPs
E) DNA polymerase–used in a polymerase chain reaction to amplify sections of DNA
SOME RESTRICTION MAPPING PRACTICE PROBLEMS
21) Construct a restriction map of a linear fragment of DNA, using the following data. Your
map should indicate the relative positions of the restriction sites along with distances from
the ends of the molecule to the restriction sites and between restriction sites:
DNA Sizes of Fragments (bp)
DNA cut with EcoRI 8000, 2000
DNA cut with BamHI
DNA cut with EcoRI + BamHI
22) Construct a restriction map of a circular DNA plasmid, using the following data. Your map 22)
should indicate the relative positions of the restriction sites along with distances between
DNA Sizes of Fragments (bp)
DNA cut with BglII 7950
DNA cut with EcoRI 7950
DNA cut with HpaI 7950
DNA cut with BglII + EcoRI 5416, 2534
DNA cut with BglII + HpaI 6632, 1318
DNA cut with EcoRI + HpaI 4098, 3852
CURRICULUM VITAE BSc PhD DSc (Med) DSc (hc) FRSE FFPH FRCPS (Glas) FRCP (Edin) FMedSci Knight's Cross of Order of Merit of Republic of Poland Fellow, National Academy of Scotland Fellow, Academy of Medical Science (United Kingdom) Honorary Member, Hungarian Academy of Science C: Users fourneraj Documents PB 2013-06 PB CV.doc CURRICULUM VITAE Name: Peter
Custom Proposal Prepared For: KeySolution Client Presented By: Proposal Date: 07/31/2013 Proposed Effective Date: 01/01/2014 Insurance Benefits Underwritten By: Fidelity Security Life Insurance Company Home Office: Kansas City, MO 64111 Premium rates are subject to change in accordance with policy provisions. A certificate will not be issued until a policy is issued in accordance with an application submitted to the underwriting company.