The following is a map of four genes on a chromosome, and the number indicates the map distances between genes:
Between which two genes would you expect the highest frequency of recombination?
A. A and W
B. E and G
C. A and E
D. A and G
E. W and E
Answer D
The frequency of crossing over between any two linked genes will be which of the following?
A. Proportional to the distance between them
B. Higher if they are recessive
C. The same as if they were not linked
D. Dependent on how many alleles there are
E. Determined by their relative dominance
Answer A
Which of the following statements describes genomic imprinting?
A. It is greatest in females because of the larger maternal contribution of cytoplasm.
B. It involves an irreversible alteration in the DNA sequence of imprinting genes.
C. Transmission of traits controlled by genomic imprinting follows Mendelian inheritance pattern
D. It explains cases in which the expression of a gene is determined by the parental origin of each allele
Answer D
The inheritance of variegated color on the leaves of certain plants (yellow patched on green leaves) was determined by the maternal parent only. What phenomenon does this describe?
A. Nuclear gene inheritance
B. Chloroplast inheritance
C. Sex-linkage
D. Genomic imprinting
E. Infectious inheritance
Answer B
Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine?
A. 31
B. 12
C. 38
D. 24
E. It cannot be determined from the information provided.
Answer B
Replication in prokaryotes differs from replication in eukaryotes for which of the following?
A. The prokaryotic chromosome has histones, whereas eukaryotic chromosomes do not.
B. Prokaryotes have telmeres, and eukaryotes do not.
C. Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many
D. The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes.
E. Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not.
Answer C
Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine (T) has been added. What would happen if a cell replicates once in the presence of this radioactive base?
A. All four bases of the DNA would be radioactive
B. DNA in both daughter cells would be radioactive
C. Radioactive thymine would pair with nonradioactive guanine
D. Neither of the two daughter cells would be radioactive
E. One of the daughter cells, but not the other, would have radioactive DNA
Answer B
Which enzyme catalyzes the elongation of DNA strand in the 5' > 3' direction?
A. topoisomerase
B. DNA ligase
C. primase
D. helicase
E. DNA polymerase III
Answer E
In an E. coli cell, a mutation in a gene prevents the helices from functioning at the origin. Which of the following would you expect as a results of this mutation?
A. No replication fork will be formed.
B. Replication will occur vis RNA polymerase alone.
C. The DNA will supercoil
D. No proofreading will occur
E. Replication will require a DNA template from another source.
Answer A
Replication of linear DNA molecule in eukaryotic cells encounters the end-replication problem. This is a consequence of which of the following?
A. Gaps left at the 5' end of the lagging strange after the primer for the last fragment is degraded
B. Gaps left at the 3' end of the lagging strand because of the need for a primer
C. DNA polymerase that cannot replicate the leading strange template to its 5' end
D. The "no ends" of a circular chromosome
E. The evolution of telomerase enzyme
Answer A
What is meant by the description "antiparallel" regarding the strands that make up DNA?
A. One strand is positively charged and the other is negatively charged.
B. Base pairings create unequal spacing between the two DNA strands.
C. The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.
D. The twisting nature o DNA creates nonparallel strands.
E. One strand contains only purines and the other contains only pyrimidines.
Answer C
Which of the following removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3' end of Okazaki fragments?
A.DNA polymerase III
B. primase
C. DNA polymerase I
D. helicase
E. ligase
Answer C
A new DNA strand elongates only in the 5' to 3' direction because
A. Okazaki fragments prevents elongation in the 3' to 5' direction
B. DNA polymerase begins adding nucleotides at the 5' end of the template
C. replication must progress toward the replication fork.
D. DNA polymerase can only add nucleotides to the free 3' end.
E. the polarity of the DNA molecule prevents addition of nucleotides at the 3' end.
Answer D
What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
A. catalyze the lengthening of telomeres
B. stabilize the unwound parental DNA
C. join Okazaki fragments together
D. unwind the parental double helix
E. synthesize RNA nucleotides to make a primer
Answer C
Which would you expect of a eukaryotic cell lacking telomerase?
A. production of Okazaki fragments
B. inability to repair thymine dimers
C. a high probability of becoming cancerous
D. high sensitivity to sunlight
E. a reduction in chromosome length following each round of DNA replication
Answer E
Why do histones bing tightly to DNA?
A. Histones are negatively charged, and DNA is positively charged.
B. Histones are highly hydrophobic, and DNA is hydrophilic
C. Both histones and DNA are strongly hydrophobic
D. Histones are positively charged, and DNA is negatively charged
E. Histones are covalently linked to the DNA
Answer D
What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
A. Polymerase can work on only one strand at a time
B. Helicases and single-strand binding proteins work at the 5' end
C. The origins of replication occur only at the 5' end
D. DNA polymerase can join new nucleotides only to the 3' end of a growing strand.
E. DNA ligase works only in the 3' > 5' direction
Answer E
Which of the following statements is true of chromatin?
A. Heterochromatin is highly condensed, whereas euchromatin is less compact
B. Euchromatin is not transcribed, whereas heterochromatin is less compact
C. Both heterochromatin and euchromatin are found in the cytoplasm
D. only euchromatin is visible under the light microscope
E. Heterochromatin is composed of DNA, whereas euchromatin is made of DNA and RNA
Answer A
Which of the following is not true of a codon?
A. It may code for the same amino acid as another codon
B. It consists of three nucleotides
C. It never codes for more than one amino acid
D. It is the basic unit of the genetic code
E. It extends from one end of a tRNA molecule
Answer E
The anticodon of a particular tRNA molecule is
A. complementary to the corresponding triplet in rRNA
B. complementary to the corresponding mRNA codon
C. the part of tRNA that bonds to a specific amino acid.
D. catalytic, making the tRNA a ribozyme
E. changeable, depending on the amino acid that attaches to the tRNA
Answer B
What polypeptide sequence will be generated based on the following mRNA codon sequence? 5' AUG-UCU-UCG-UUA-UCC-UUG 3'
A. met-ser-leu-ser-leu-ser
B. met-arg-glu-arg-glu-arg
C. met-leu-phe-arg-glu-glu
D. met-glu-arg-arg-gln-leu
E. met-ser-ser-leu-ser-leu
Answer E
Which of the following statements best describes the termination of transcription in prokaryotes?
A. RNA polymerase transcribes though the terminator sequence, causing the polymerase to fall off the DNA and release the transcript
B. RNA polymerase transcribes through an intron, and the polymerase lets go of the transcript.
C. Once transcription has initiated, RNA polymerase transcribes until it reaches the end of the chromosomes
D. RNA polymerase transcribes though the polyadenylation signal, causing proteins to associate with the transcript and cut it free from the polymerase
E. RNA polymerase transcribes through a stop codon, causing the polymerase to stop advancing through the gene and release the mRNA
Answer A
Transcription initiation in eukaryotes requires which of the following in addition to RNA polymerase?
A. start and stop codons
B. ribosomes and tRNA
C. aminoacyl synthetase
D. several transcription factor (TFs)
E. the protein product of the promoter
Answer D
Which of the following help(s) to stabilize mRNA by inhibiting its degradation?
A. RNA polymerase
B. TATA box
C. 5' cap and pol (A) tail
D. introns
E. exons
Answer C
Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that includes
A. fusion with other newly transcribed mRNA
B. Linkage to histone molecules
C. excision of introns
D. fusion into circular forms
E. union with ribosomes
Answer C
Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the
A. binding of ribosomes to mRNA
B. shape of the A and P sites of ribosomes
C. attachment of amino acids to tRNAs
D. bonding of the anticodon to the codon
E. both C and D
Answer E
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.
The dipeptide that will form will be
A. proline-threonin
B. alanine-alanine
C. theonine-glycine
D. cysteine-alanine
E. glycine-cysteine
Answer A
The tRNA shown above has its 3' end projecting beyond its 5' end. What will occur at this 3' end?
A. The 5' cap of the mRNA will become covalently bound.
B. The amino acid binds covalently
C. The small and large subunits of the ribosome will attach to it.
D. The codon and anticodon complement one another
E. The excess nucleotides (ACCA) will be cleaved off at the ribosome
Answer B
There are 61 codons that specify 20 amino acids, but only 45 tRNAs. This is best explained by the fact that
A. the rules for base pairing between the third base of a codon and anticodon on tRNA are flexible.
B. many codons are never used, so the tRNAs that recognize them are dispensable.
C. the DNA codes for all 61 tRNAs but some are then destroyed
D. some tRNAs have anticodons that recognize four or more different codons
E. competitive exclusion forces some tRNAs to be destroyed by nucleases
Answer A
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 E site, and the A site remains vacant
4. A small ribosomal subunit binds with mRNA
5. Ribosome and tRNAs translocate one codon position
A. 2,4,5,1,3
B. 1,3,2,4,5
C. 5,4,3,2,1
D. 4,1,3,2,5
E. 4,1,2,5,3
Answer E
The process of translation, whether in prokaryotes or eukaryotes, requires mRNAs, charged tRNAs, ribosomal subunits, and which of the following?
A. additional protein factors for initiation, elongation and termination
B. polymerases plus GTP
C. signal polypeptides plus release factor
D. SRP plus chaperones
Answer A
What does the operon model attempt to explain?
A. bacterial resistance to antibiotics
B. how genes move between homologous regions of DNA
C. the mechanism of viral attachment to a host cell
D. horizontal transmission of plant viruses
E. the coordinated control of gene expression in bacteria
Answer E
The tryptophan operon is a repressible operon that is
A. turned on only when glucose is present in the growth medium
B. turned off whenever tryptophan is added to the growth medium
C. turn on only hone tryptophan is present in the growth medium
D. turned off only when glucose is present in the growth medium
E. permanently turned on
Answer B
A mutation in this section of DNA could influence the binding of RNA polymerase to the DNA:
A. corepressor
B. inducer
C. repressor
D. promoter
E. operon
Answer D
Allolactose induces the synthesis of the enzyme B-galactose(LacZ) that breaks down lactose. An E. coli cell is presented for the first time with the sugar lactose ( containing allolactose) as a potential food source. Which of the following occurs when the lactose enters the cell?
A. RNA polymerase attaches to the regulator
B. The repressor protein attaches to the regulator
C. Allolactose binds to the repressor protein and inactivated it
D. Allolactose binds to the regulator gene
E. The repressor protein and allocates bind to RNA polymerase
Answer C
Muscle cells and nerve cells in one species of animal owe their differences in structure to
A. having different chromosomes
B. using different genetic codes
C. having different genes
D. having unique ribosomes
E. having different genes expressed.
Answer E
Suppose a researcher becomes proficient with a technique that allows him to move DNA segments within a prokaryotic genome. If he moves the operator for the Lac operon to the far end of the operon (past LacA gene), which of the following would likely occur in the cell with the modified Lac operon?
A. The Lac structural genes will be transcribed continuously, even when lactose is absent
B. The inducer will no longer bind to the repressor
C. The repressor protein will no longer be produced
D. The repressor will no longer bind to the operator
E. The operon will never be transcribed
Answer A
Two potential mechanisms that eukaryotic cells use to regulate transcription are
A. DNA amplification and histone methylation
B. DNA acetylation and methylation
C. DNA methylation and histone amplification
D. DNA methylation and histone modifcations
E. histone amplification and DNA acetylation
Answer D
This protein molecule binds to a control element far from the promoter to stimulate transcription.
A. terminator
B. promoter
C. Enhancer
D. repressor
E. activator
Answer E
The phenomenon in which mRNA molecules in a cell are destroyed if they have a sequence complementary to an introduced double-stranded RNA is called
A. RNA disposal
B. RNA obstruction
C. RNA interference
D. RNA blocking
E. RNA targeting
Answer C
If a particular operon encodes enzymes for making an essential amino acid and is regulated like the trip operon, then the
A. amino acid acts as a corepressor
B. enzymes produced are called inducible enzymes
C. amino acid inactivates the repressor
D. repressor is active in the absence of the amino acid
E. amino acid turns on transcription of the operon
Answer A
The functioning of enhancers is an example of
A. post-trasnlational control that activates certain proteins
B. a post-transcriptional mechanism for editing mRNA
C. transcriptional control of gene expression
D. a eukaryotic equivalent of prokaryotic terminator functioning.
E. the stimulation of translation by initiation factors
Answer C
Which of the following mechanism is (are) used to coordinate the expression of multiple, related genes in eukaryotic cells?
A. Environmental signals enter the cell and bind directly to promoters
B. The genes share common enhancer, which allows appropriate actuators to turn on their transcription at the same time.
C. The genes are organized into a large operon, allowing them to be coordinately controlled as a single unit
D. A single repressor is able to turn off several related genes.
Answer B