AP Unit 7: Molecular Genetics Practice Test	Name:__________________________________

____	1)	In an analysis of the nucleotide composition of DNA, which of the following will be found?
A)	A = C	C)	A + C = G + T	
B)	A = G and C = T	D)	G + C = T + A	

____	2)	Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons?
A)	Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not.	
B)	Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many.	
C)	The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes.	
D)	Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not.	
E)	Prokaryotes have telomeres, and eukaryotes do not.	

____	3)	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)	No proofreading will occur.	
B)	No replication fork will be formed.	
C)	The DNA will supercoil.	
D)	Replication will occur via RNA polymerase alone.	
E)	Replication will require a DNA template from another source.	

____	4)	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 causes linear ends of the newly replicated DNA to circularize.	
D)	It catalyzes the lengthening of telomeres, compensating for the shortening that occurs during replication.	
E)	It adds numerous GC pairs which resist hydrolysis and maintain chromosome integrity.	

____	5)	At a specific area of a chromosome, the sequence of nucleotides below is present where the chain opens to form a replication fork:
3' C C T A G G C T G C A A T C C 5'
An RNA primer is formed starting at the underlined T ( T) of the template. Which of the following represents the primer sequence?
A)	5' G C C T A G G 3'	C)	5' A C G T T A G G 3'	E)	5' G C C U A G G 3'	
B)	3' G C C T A G G 5'	D)	5' A C G U U A G G 3'	

____	6)	What is the function of DNA polymerase III?
A)	to unwind the DNA helix during replication	
B)	to seal together the broken ends of DNA strands	
C)	to add nucleotides to the 3' end of a growing DNA strand	
D)	to degrade damaged DNA molecules	
E)	to rejoin the two DNA strands (one new and one old) after replication	

____	7)	What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
A)	It synthesizes RNA nucleotides to make a primer.	
B)	It catalyzes the lengthening of telomeres.	
C)	It joins Okazaki fragments together.	
D)	It unwinds the parental double helix.	
E)	It stabilizes the unwound parental DNA.	

____	8)	In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (15N) and then transferred them to a medium containing 14N. Which of the results in the figure to the right would be expected after one round of DNA replication in the presence of 14N?
A)	A	B)	B	C)	C	D)	D	E)	E	

____	9)	What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
A)	The origins of replication occur only at the 5' end.	
B)	Helicases and single–strand binding proteins work at the 5' end.	
C)	DNA polymerase can join new nucleotides only to the 3' end of a growing strand.	
D)	DNA ligase works only in the 3'  5' direction.	
E)	Polymerase can work on only one strand at a time.	

____	10)	Which of the following statements best describes the termination of transcription in prokaryotes?
A)	RNA polymerase transcribes through the polyadenylation signal, causing proteins to associate with the transcript and cut it free from the polymerase.	
B)	RNA polymerase transcribes through the terminator sequence, causing the polymerase to separate from the DNA and release the transcript.	
C)	RNA polymerase transcribes through an intron, and the snRNPs cause the polymerase to let go of the transcript.	
D)	Once transcription has initiated, RNA polymerase transcribes until it reaches the end of the chromosome.	
E)	RNA polymerase transcribes through a stop codon, causing the polymerase to stop advancing through the gene and release the mRNA.	

____	11)	A eukaryotic transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a protein consisting of approximately 400 amino acids. This is best explained by the fact that
A)	many noncoding stretches of nucleotides are present in eukaryotic DNA.	
B)	there is redundancy and ambiguity in the genetic code.	
C)	many nucleotides are needed to code for each amino acid.	
D)	nucleotides break off and are lost during the transcription process.	
E)	there are termination exons near the beginning of mRNA.	

____	12)	Alternative RNA splicing
A)	is a mechanism for increasing the rate of transcription.	
B)	can allow the production of proteins of different sizes from a single mRNA.	
C)	can allow the production of similar proteins from different RNAs.	
D)	increases the rate of transcription.	
E)	is due to the presence or absence of particular snRNPs.	

____	13)	There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best explained by the fact that
A)	some tRNAs have anticodons that recognize four or more different codons.	
B)	the rules for base pairing between the third base of a codon and tRNA are flexible.	
C)	many codons are never used, so the tRNAs that recognize them are dispensable.	
D)	the DNA codes for all 61 tRNAs but some are then destroyed.	
E)	competitive exclusion forces some tRNAs to be destroyed by nucleases.	

____	14)	When the ribosome reaches a stop codon on the mRNA, no corresponding tRNA enters the A site. If you could take a snapshot of the translation reaction at this point, which of the following would you see?
A)	an assembled ribosome with a polypeptide attached to the tRNA in the P site	
B)	separated ribosomal subunits, a polypeptide, and free tRNA	
C)	an assembled ribosome with a separated polypeptide	
D)	separated ribosomal subunits with a polypeptide attached to the tRNA	
E)	a cell with fewer ribosomes	

____	15)	What is the effect of a nonsense mutation in a gene?
A)	It changes an amino acid in the encoded protein.	
B)	It has no effect on the amino acid sequence of the encoded protein.	
C)	It introduces a premature stop codon into the mRNA.	
D)	It alters the reading frame of the mRNA.	
E)	It prevents introns from being excised.	

____	16)	Which of the following DNA mutations is the most likely to be damaging to the protein it specifies?
A)	a base–pair deletion	D)	a point mutation	
B)	a codon substitution	E)	a substitution in the last base of a codon	
C)	a codon deletion	

____	17)	A possible sequence of nucleotides in the template strand of DNA that would code for the polypeptide sequence phe–leu–ile–val would be
A)	5' TTG–CTA–CAG–TAG 3'.	D)	3' AAA–AAT–ATA–ACA 5'.	
B)	3' AAC–GAC–GUC–AUA 5'.	E)	3' AAA–GAA–TAA–CAA 5'.	
C)	5' AUG–CTG–CAG–TAT 3'.	

A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The following charged transfer RNA molecules (with their anticodons shown in the 3' to 5' direction) are available. Two of them can correctly match the mRNA so that a dipeptide can form.

____	18)	The dipeptide that will form will be
A)	cysteine–alanine.	C)	glycine–cysteine.	E)	threonine–glycine.	
B)	proline–threonine.	D)	alanine–alanine.	

____	19)	A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the growing polypeptide on the other tRNA (#2) in the ribosome already.  Where does tRNA #1 move to after this bonding of lysine to the polypeptide?
A)	A site	B)	P site	C)	E site	D)	B site	E)	directly to the cytosol	

____	20)	In eukaryotic cells, transcription cannot begin until
A)	the two DNA strands have completely separated and exposed the promoter.	
B)	several transcription factors have bound to the promoter.	
C)	the 5' caps are removed from the mRNA.	
D)	the DNA introns are removed from the template.	
E)	DNA nucleases have isolated the transcription unit.	

____	21)	The anticodon of a particular tRNA molecule is
A)	complementary to the corresponding mRNA codon.	
B)	complementary to the corresponding triplet in rRNA.	
C)	the part of tRNA that bonds to a specific amino acid.	
D)	changeable, depending on the amino acid that attaches to the tRNA.	
E)	catalytic, making the tRNA a ribozyme.	

____	22)	The tryptophan operon is a repressible operon that is
A)	permanently turned on.	
B)	turned on only when tryptophan is present in the growth medium.	
C)	turned off only when glucose is present in the growth medium.	
D)	turned on only when glucose is present in the growth medium.	
E)	turned off whenever tryptophan is added to the growth medium.	

____	23)	Which of the following is a protein produced by a regulatory gene?
A)	operon	B)	inducer	C)	promoter	D)	repressor	E)	corepressor	

____	24)	For a repressible operon to be transcribed, which of the following must occur?
A)	A corepressor must be present.	
B)	RNA polymerase and the active repressor must be present.	
C)	RNA polymerase must bind to the promoter, and the repressor must be inactive.	
D)	RNA polymerase cannot be present, and the repressor must be inactive.	
E)	RNA polymerase must not occupy the promoter, and the repressor must be inactive.	

____	25)	The phenomenon in which RNA molecules in a cell are destroyed if they have a sequence complementary to an introduced double-stranded RNA is called
A)	RNA interference.	C)	RNA blocking.	E)	RNA disposal.	
B)	RNA methylation.	D)	RNA targeting.	
____	26)	At the beginning of this century there was a general announcement regarding the sequencing of the human genome and the genomes of many other multicellular eukaryotes. There was surprise expressed by many that the number of protein-coding sequences was much smaller than they had expected. Which of the following could account for most of the rest?
A)	"junk" DNA that serves no possible purpose	
B)	rRNA and tRNA coding sequences	
C)	DNA that is translated directly without being transcribed	
D)	non-protein-coding DNA that is transcribed into several kinds of small RNAs with biological function	
E)	non-protein-coding DNA that is transcribed into several kinds of small RNAs without biological function	

____	27)	Which of the following accounts for someone who has had a herpesvirus–mediated cold sore or genital sore getting flare–ups for the rest of his or her life?
A)	re–infection by a closely related herpesvirus of a different strain	
B)	re–infection by the same herpesvirus strain	
C)	co–infection with an unrelated virus that causes the same symptoms	
D)	copies of the herpesvirus genome permanently maintained in host nuclei	
E)	copies of the herpesvirus genome permanently maintained in host cell cytoplasm	

____	28)	Which of the following statements describes the lysogenic cycle of lambda (λ) phage?
A)	After infection, the viral genes immediately turn the host cell into a lambda–producing factory, and the host cell then lyses.	
B)	Most of the prophage genes are activated by the product of a particular prophage gene.	
C)	The phage genome replicates along with the host genome.	
D)	Certain environmental triggers can cause the phage to exit the host genome, switching from the lytic to the lysogenic.	
E)	The phage DNA is incorporated by crossing over into any nonspecific site on the host cell's DNA.	

____	29)	What is the function of reverse transcriptase in retroviruses?
A)	It hydrolyzes the host cell's DNA.	
B)	It uses viral RNA as a template for DNA synthesis.	
C)	It converts host cell RNA into viral DNA.	
D)	It translates viral RNA into proteins.	
E)	It uses viral RNA as a template for making complementary RNA strands.	

____	30)	Assume that you are trying to insert a gene into a plasmid. Someone gives you a preparation of genomic DNA that has been cut with restriction enzyme X. The gene you wish to insert has sites on both ends for cutting by restriction enzyme Y. You have a plasmid with a single site for Y, but not for X. Your strategy should be to
A)	insert the fragments cut with restriction enzyme X directly into the plasmid without cutting the plasmid.	
B)	cut the plasmid with restriction enzyme X and insert the fragments cut with restriction enzyme Y into the plasmid.	
C)	cut the DNA again with restriction enzyme Y and insert these fragments into the plasmid cut with the same enzyme.	
D)	cut the plasmid twice with restriction enzyme Y and ligate the two fragments onto the ends of the DNA fragments cut with restriction enzyme X.	
E)	cut the plasmid with restriction enzyme X and then insert the gene into the plasmid.	

____	31)	Which enzyme was used to produce the molecule in Figure 20.1?
A)	ligase	C)	a restriction enzyme	E)	DNA polymerase	
B)	transcriptase	D)	RNA polymerase	

____	32)	
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?

A)		C)		
B)		D)		
E)		

____	33)	How do we describe transformation in bacteria?
A)	the creation of a strand of DNA from an RNA molecule	
B)	the creation of a strand of RNA from a DNA molecule	
C)	the infection of cells by a phage DNA molecule	
D)	the type of semiconservative replication shown by DNA	
E)	assimilation of external DNA into a cell	

Use the following information to answer questions 34 and 35.

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.

____	34)	Bacteria that contain the plasmid, but not the eukaryotic gene, would grow
A)	in the nutrient broth plus ampicillin, but not in the broth containing tetracycline.	
B)	only in the broth containing both antibiotics.	
C)	in the broth containing tetracycline, but not in the broth containing ampicillin.	
D)	in all four types of broth.	
E)	in the nutrient broth without antibiotics only.	

____	35)	Bacteria that do not take up any plasmids would grow on which media?
A)	the nutrient broth only	
B)	the nutrient broth and the tetracycline broth	
C)	the nutrient broth and the ampicillin broth	
D)	the tetracycline broth and the ampicillin broth	
E)	all three broths	

____	36)	A paleontologist has recovered a bit of tissue from the 400 -year-old preserved skin of an extinct dodo (a bird). To compare a specific region of the DNA from the sample with DNA from living birds, which of the following would be most useful for increasing the amount of dodo DNA available for testing?
A)	RFLP analysis	D)	gel electrophoresis	
B)	polymerase chain reaction (PCR)	E)	Southern blotting	
C)	electroporation	

AP Unit 7: Molecular Genetics Practice Test
Answer Section

MULTIPLE CHOICE

	1)	ANS:	C	TOP:	Concept 16.1		

	2)	ANS:	B	TOP:	Concept 16.2		

	3)	ANS:	B	TOP:	Concept 16.2		

	4)	ANS:	D	TOP:	Concept 16.2		

	5)	ANS:	D	TOP:	Concept 16.2		

	6)	ANS:	C	TOP:	Concept 16.2		

	7)	ANS:	C	TOP:	Concept 16.2		

	8)	ANS:	D	TOP:	Concept 16.2		

	9)	ANS:	C	TOP:	Concept 16.2	

	10)	ANS:	B	TOP:	Concept 17.2		

	11)	ANS:	A	TOP:	Concept 17.3		

	12)	ANS:	B	TOP:	Concept 17.3		

	13)	ANS:	B	TOP:	Concept 17.4		

	14)	ANS:	A	TOP:	Concept 17.4		

	15)	ANS:	C	TOP:	Concept 17.5		

	16)	ANS:	A	TOP:	Concept 17.5		

	17)	ANS:	E	TOP:	Concept 17.1		

	18)	ANS:	B	TOP:	Concept 17.4		

	19)	ANS:	B	TOP:	Concept 17.4		

	20)	ANS:	B	TOP:	Concept 17.2

	21)	ANS:	A	TOP:	Concept 17.4	

	22)	ANS:	E	TOP:	Concept 18.1		

	23)	ANS:	D	TOP:	Concept 18.1		

	24)	ANS:	C	TOP:	Concept 18.1		

	25)	ANS:	A	TOP:	Concept 18.3		

	26)	ANS:	D	TOP:	Concept 18.3		

	27)	ANS:	D	TOP:	Concept 19.2		

	28)	ANS:	C	TOP:	Concept 19.2		

	29)	ANS:	B	TOP:	Concept 19.3		

	30)	ANS:	C	TOP:	Concept 20.1		

	31)	ANS:	C	TOP:	Concept 20.1		

	32)	ANS:	B	TOP:	Concept 20.2		

	33)	ANS:	E	TOP:	Concept 16.1		

	34)	ANS:	D	TOP:	Concept 20.1		

	35)	ANS:	A	TOP:	Concept 20.1		

	36)	ANS:	B	TOP:	Concept 20.2