Genetics Final Exam Quiz
Do you go to Westmont College and are taking human anatomy class? If so, then I have prepared a genetics quiz that is designed to help you review what we covered this semester. Do give it a try and see just how much you know about it as you prepare for the final exam. All the best and keep on revising!
- 1.
In our experiment in lab, which bacteria was able to grow on the LB + amp plates?
- A.
All the bacteria, regardless of whether that had the plasmid or not.
- B.
Only the bacteria that contained the plasmid.
- C.
Only the bacteria that lacked the plasmid.
Correct Answer
B. Only the bacteria that contained the plasmid. -
- 2.
Hershey and Chase's experiment helped prove that DNA is the genetic material. What if they had used a phage that had RNA as its genetic material? Would their experimental design still work?
- A.
Yes, because their use of RNase would tell them whether RNA was critical.
- B.
No, because the phage RNA would incorporate both [32P] and [35S].
- C.
No, because the phage RNA would NOT incorporate either [32P] or [35S].
- D.
Yes, because, like DNA, the phage RNA would incorportate [32P] and not [35S]
Correct Answer
D. Yes, because, like DNA, the phage RNA would incorportate [32P] and not [35S] -
- 3.
Why did Chargaff’s data support the idea that DNA was the genetic material while Levene’s did not? Choose one answer.
- A.
A. Levene’s data implied the nucleotides were randomly assembled in cells but Chargaff proved they were in a specific order.
- B.
B. Chargaff proved that each species had different values for each base (although within species A=T & G=C), suggesting that DNA was different between them.
- C.
C. Levene’s data showed there wasn’t enough DNA in cells to encode all the necessary information but Chargaff disproved this.
- D.
D. Levene found that RNA & DNA used the exact same nucleotides while Chargaff proved that they didn’t.
Correct Answer
B. B. Chargaff proved that each species had different values for each base (although within species A=T & G=C), suggesting that DNA was different between them. -
- 4.
In Griffith's experiments, why did the rat die when it was injected with live R strain and heat-killed S strain (choose most appropriate answer)?
- A.
Some of the S strain bacteria came back to live after acquiring some fragments of the R strain.
- B.
One R strain bacterium must have spontaneously mutated to become virulent.
- C.
Some of the R strain bacteria took in a fragment of the S strain DNA that encoded for synthesis of the polysaccharide coat.
- D.
Some of the R strain bacteria took in a fragment of the S strain DNA that encoded for ampicillin resistance.
- E.
The R strain surrounded itself with some of the polysaccharide coating from the dead S strain.
Correct Answer
C. Some of the R strain bacteria took in a fragment of the S strain DNA that encoded for synthesis of the polysaccharide coat. -
- 5.
What is NOT a required property of genetic material?
- A.
Reliable storage of all necessary information for life
- B.
Ability to be replicated
- C.
Unchanging over time
- D.
Controlled expression of information
- E.
All are required properties of genetic material
Correct Answer
C. Unchanging over time -
- 6.
Who isolated DNA from human cells for the first time?
- A.
Schwann and Schleiden
- B.
Watson and Crick
- C.
Meischer
- D.
Griffith
- E.
Mendel
Correct Answer
C. Meischer -
- 7.
One strand of the DNA helix is 3’-GCCTATTTGACT-5’. What is the sequence and direction of its complementary strand?
- A.
5’-TCAGTTTATCCG-3’
- B.
B. 3’-CGGAUAAACUGA-5’
- C.
C. 3’-TCAGTTTATCCG-5’
- D.
D. 5’-CGGATAAACTGA-3’
- E.
E. 5’-CGGAUAAACUGA-3’
- F.
F. 3’-CGGATAAACTGA-5’
Correct Answer
D. D. 5’-CGGATAAACTGA-3’ -
- 8.
In a nucleotide, the phosphate is attached to which carbon of the pentose sugar?
- A.
3'
- B.
2'
- C.
4'
- D.
5'
- E.
1'
Correct Answer
D. 5' -
- 9.
If the conservative model of DNA replication WAS correct, what bands would Meselson and Stahl have seen after the 2nd generation of E. coli growth?
- A.
One, thick intermediate band
- B.
Equal quantities of an intermediate band and a heavy band
- C.
Large, thick “light” band and a thinner “heavy” band
- D.
Thinner “light” band and a large, thick “heavy” band
- E.
Light, intermediate, and heavy bands
Correct Answer
C. Large, thick “light” band and a thinner “heavy” band -
- 10.
DNA polymerase I adds nucleotides to the
- A.
5' end of the primer.
- B.
3' end of the primer.
- C.
5’ end of the new DNA strand after the primer is removed
- D.
3’ end of the new DNA strand after the primer is removed
Correct Answer
D. 3’ end of the new DNA strand after the primer is removed -
- 11.
Why did Arthur Kornberg find DNA pol I and not DNA pol III even though the latter is the enzyme responsible for the majority of DNA replication?
- A.
There is significantly more DNA pol I in each cell.
- B.
E. coli don't have DNA pol III.
- C.
DNA pol III does not work in a cell-free, in vitro DNA synthesis system.
- D.
He did find DNA pol III but called it the Kornberg enzyme.
Correct Answer
A. There is significantly more DNA pol I in each cell. -
- 12.
The discontinuous aspect of DNA replication on the lagging strand is caused by
- A.
The 5' to 3' polarity restriction in polymerization.
- B.
Random priming
- C.
Polymerase slippage
- D.
Discontinuous helicase activity.
- E.
None of the above
Correct Answer
A. The 5' to 3' polarity restriction in polymerization. -
- 13.
What technique would you use if you wanted to determine which chromosome contained a certain gene?
- A.
In situ hybridization
- B.
Southern blotting
- C.
Transformation
- D.
FISH
- E.
Restriction mapping
Correct Answer
D. FISH -
- 14.
The origin of replication in a prokaryotic chromosome has a lot of A's and T's. Why does this make sense?
- A.
There is no reason. It is pure chance but has been conserved over time.
- B.
A's and T's are more easily recognized by the proteins of the replisome
- C.
They make up the restriction site recognized by the restriction endonuclease
- D.
They are easier to pry apart (i.e. require less energy).
Correct Answer
D. They are easier to pry apart (i.e. require less energy). -
- 15.
An absorbance value of 1 at 260nm for DNA indicates the sample contains 50 micrograms/milliliter of DNA. In contrast, an absorbance value of 1 at 260 nm for RNA indicates the sample contains 40 micrograms/milliliter of RNA. Why the difference?
- A.
DNA is always heavier because of the de-oxyribose in the nucleotides
- B.
RNA absorbs more UV light because it is single-stranded
- C.
Uracil absorbs more UV light because it contains more nitrogen.
- D.
RNA is less-stable so degrades more easily.
Correct Answer
B. RNA absorbs more UV light because it is single-stranded -
- 16.
How does RNA differ from DNA?
- A.
It forms left-handed helix.
- B.
It uses only purines
- C.
In RNA, the amount of G's may not equal the amount of C's
- D.
It is not negatively charged
- E.
None of the above
Correct Answer
C. In RNA, the amount of G's may not equal the amount of C's -
- 17.
Zamecnick mixed together 3 components in his "cell-free in vitro protein synthesis system." In which component were the ribosomes that he discovered?
- A.
The radioactive amino acids
- B.
The non-radioactive amino acids
- C.
The E. coli lysate
- D.
The ATP regeneration system
- E.
The Rat Liver Extract
Correct Answer
E. The Rat Liver Extract -
- 18.
Transcription in prokaryotes terminates at
- A.
A gene's stop codon
- B.
A region beyond the end of the RNA-coding region
- C.
A region called the TATA box
- D.
A region called the promoter
Correct Answer
B. A region beyond the end of the RNA-coding region -
- 19.
What do homologous chromosomes have in common?
- A.
They are the same size.
- B.
They contain the same genes.
- C.
They will show the same banding pattern when stained
- D.
Their centromere is in the same location.
- E.
All of the above.
Correct Answer
E. All of the above. -
- 20.
Chromatin remodeling affects
- A.
G-banding pattern of chromosomes
- B.
Gene activity (how often a gene is "on" and "off")
- C.
MRNA processing
- D.
Rate of DNA replication
- E.
Number of nucleotides in a genome
Correct Answer
B. Gene activity (how often a gene is "on" and "off") -
- 21.
What property is true about telomeras?
- A.
Polymerizes DNA from a DNA template
- B.
Polymerizes DNA from an RNA template
- C.
Polymerizes RNA from an RNA template
- D.
Polymerizes RNA from a DNA template
- E.
Requires no template to polymerize nucleic acid
Correct Answer
B. Polymerizes DNA from an RNA template -
- 22.
Which answer below would be a reasonable size for a prokaryotic chromosome?
- A.
54,200 bp
- B.
275,000 bp
- C.
1.39 million bp
- D.
4.01 billion bp
- E.
2.39 trillion bp
Correct Answer
C. 1.39 million bp -
- 23.
Supercoiled DNA
- A.
Have more histones in their chromain
- B.
May have slightly more or less than 10.4 base-pairs per helical turn
- C.
Is only found in mitotic chromosomes
- D.
Is never found naturally in cells.
- E.
Forms a left-handed helix
Correct Answer
B. May have slightly more or less than 10.4 base-pairs per helical turn -
- 24.
The function of the sigma factor in a prokaryote is:
- A.
To prime the mRNA
- B.
To unwind the DNA helix during replication
- C.
To facilitate elongation
- D.
To bind strongly to the promoter
- E.
To mediate termination
Correct Answer
D. To bind strongly to the promoter -
- 25.
Why are frameshifts one of the most severe types of mutations?
- A.
They occur only in gametes
- B.
More than one chromosome is affected.
- C.
They change a nucleotide's base-pairing, causing it to base-pair with a different nucleotide
- D.
They not only affect one codon but all codons downstream.
Correct Answer
D. They not only affect one codon but all codons downstream. -
- 26.
What fluctuated in the fluctuation experiment of Luria and Delbruck?
- A.
The number of resistant cells in different samplings from the same flask.
- B.
The number of resistant cells from different independent cultures.
- C.
The ratio of phages to bacteria (or multiplicity of infection) in the experiment.
- D.
The probability of a bacterium coming in contact with a phage.
- E.
The mutation rate in different bacteria.
Correct Answer
B. The number of resistant cells from different independent cultures. -
- 27.
Tandem repeats are considered mutation hotspots because
- A.
They contain a lot of methylated bytosines
- B.
They are more susceptible to deamination.
- C.
They do not contain genes and therefore have no selective pressure to remain free of mutation.
- D.
They are common locations of DNA polymerase slippage.
- E.
None of the above; they are not considered mutation hotspots..
Correct Answer
D. They are common locations of DNA polymerase slippage. -
- 28.
In Garrod's understanding, why did his patients have AKU?
- A.
Their mother ate too much tyrosine while she was pregnant.
- B.
Their mother ate too little tyrosine while she was pregnant.
- C.
They inherited some defect in a metabolic pathway from their parents.
- D.
They were exposed to a mutation-causing toxin.
- E.
He never knew why.
Correct Answer
C. They inherited some defect in a metabolic pathway from their parents. -
- 29.
A mutation that deaminates one of teh bases would be considered a
- A.
Base analog
- B.
Intercalating agent.
- C.
Base-modifying agent.
- D.
Methylating agent.
- E.
None of the above; it is not a mutagen at all.
Correct Answer
C. Base-modifying agent. -
- 30.
A mutation is induced in Neurospora that causes the loss of function of an enzyme essential to the biosynthesis of the vitamin pyridoxine. On which medium will such organisms NOT grow?
- A.
Minimal
- B.
Minimal + all vitamins
- C.
Complete
- D.
Minimal + pyridoxine
Correct Answer
A. Minimal -
- 31.
Occasionally, a mutation is discovered that suppresses termination and allows a stop codon to be read as an amino acid without changing the codon. What has this mutation most likely done?
- A.
Changed the anticodon sequence for an amino-acid carrying tRNA.
- B.
Changed the ribosome binding capacity.
- C.
Altered a terminating tRNA so that it carries an amino acid.
- D.
Changed the termination factor.
Correct Answer
A. Changed the anticodon sequence for an amino-acid carrying tRNA. -
- 32.
Based on the genetic code found in your textbook, which peptide would this mRNA encode in a prokaryote: 5' UUAUGCCAAGGUUGUAAG 3'?
- A.
FMet-Pro-Arg-Leu
- B.
Met-Pro-Arg-Leu
- C.
FMet-Leu-Cys-Gln-Gly-Cys-Lys
- D.
Leu-Cys-Gln-Gly-Cys-Lys
Correct Answer
A. FMet-Pro-Arg-Leu -
- 33.
The role of tautomerism in causing mutations relates to the fact that the process ultimately affects
- A.
The covalent bonding between the bases
- B.
Complementary base pairing
- C.
The deoxyribose sugar
- D.
The ability of DNA to replicate at all
- E.
The phosphate group
Correct Answer
B. Complementary base pairing -
- 34.
In eukaryotes, pyrimidine dimers are usually fixed by
- A.
Transposons
- B.
Photoreactivation
- C.
Base excision repair
- D.
Nucleotide excision repair
- E.
Mismatch repair
Correct Answer
D. Nucleotide excision repair -
- 35.
Imagine that Barbara McClintock found a speckled kernel of maize that did lack the Activator (Ac). Which is the most plausible explanation?
- A.
Ds had a reverse mutation in its transposase to regain functionality.
- B.
The maize was exposed to a chemical mutagen.
- C.
It must have lacked Ds as well.
- D.
She was wrong; it must have still had Ac. No other explanation is plausible.
Correct Answer
A. Ds had a reverse mutation in its transposase to regain functionality. -
- 36.
CAP is a trans-acting factor that binds the _______________ of the lac operon and ___________________ the rate of transcription initiation.
- A.
LacZ gene; decreases
- B.
LacZ gene; increases
- C.
Operator; decreases
- D.
Operator; increases
- E.
Promoter; decreases
- F.
Promoter; increases
Correct Answer
F. Promoter; increases -
- 37.
Which enzyme is required for insertion of new retrotransposons?
- A.
Transposase
- B.
Retrotransposase
- C.
Reverse transcriptase
- D.
Reverse transposase
- E.
Reverse polymerase
Correct Answer
C. Reverse transcriptase -
- 38.
In mismatch repair in prokaryotes, the original non-mutated parent strand is identifiable because
- A.
It is methylated.
- B.
It is unmethylated.
- C.
Its sugar-phosphate backbone is not distorted.
- D.
There are extra nucleotides in it.
- E.
There are missing nucleotides in it.
Correct Answer
A. It is methylated. -
- 39.
The tryptophan operon contains genes that encode enzymes used to synthesize the amino acid, tryptophan (trp). The bacterium only wants it turned on if it has run out of teh amino acid and cannot find it in the environment. The trp repressor is a DNA binding protein that is involved in the negative regulation of this operon. Its inducer is the trp amino acid itself. What is likely true?
- A.
The binding of trp to the trp repressor causes it to decrease its affinity for DNA.
- B.
The binding of trp to the trp repressor causes it to increase its affinity for DNA.
- C.
The presence of trp induces the expression of the gene encoding the trp repressor
- D.
The presence of trp turns off the expression of the gene encoding the trp repressor.
Correct Answer
B. The binding of trp to the trp repressor causes it to increase its affinity for DNA. -
- 40.
Genes that are continually transcribed are called:
- A.
Transposons.
- B.
Operator genes.
- C.
Constitutive genes.
- D.
Operons.
- E.
Repressor genes.
- F.
Non-inducible genes.
Correct Answer
C. Constitutive genes. -
- 41.
By definition, transgenic organisms
- A.
Have an extra gene, usually from a different species.
- B.
Have had their genomes annotated.
- C.
Have been born in a laboratory through in vitro fertilization.
- D.
Lack a gene.
- E.
Have had their genomes sequenced.
Correct Answer
A. Have an extra gene, usually from a different species. -
- 42.
The sister chromatids separate in which stage of meiosis?
- A.
Telophase I
- B.
Telophase II
- C.
Anaphase I
- D.
Anaphase II
- E.
Prophase I
Correct Answer
D. Anaphase II -
- 43.
Which is an example of a sequence polymorphism?
- A.
EST
- B.
STS
- C.
SNP
- D.
ORF
- E.
Contig
Correct Answer
C. SNP -
- 44.
In a eukaryotic organism with 28 chromosomes in one of its somatic cells, what would be the number of chromosomes present in each of its daughter cells after 2 rounds of mitosis
- A.
12
- B.
56
- C.
28
- D.
112
- E.
6
Correct Answer
C. 28 -
- 45.
The first genome to be sequenced was from a
- A.
Transposon
- B.
Fruit fly
- C.
Phage or virus
- D.
Human
- E.
Type of prokaryote
Correct Answer
C. Phage or virus -
- 46.
The most common method of creating a genetically modified plant involves using a
- A.
Retrovirus
- B.
Phage
- C.
Transposon
- D.
Parasitic bacterium
- E.
Retrotransposons
Correct Answer
D. Parasitic bacterium -
- 47.
Roundup-Ready crops are resistant to
- A.
A parasitic bacteria
- B.
High salinity
- C.
Fungus
- D.
An herbicide
- E.
Insects
Correct Answer
D. An herbicide -
- 48.
An F1 female crossed with a male with all recessive traits is called a ________________ cross.
- A.
F1
- B.
F2
- C.
Parental
- D.
Test
Correct Answer
D. Test -
- 49.
Preimplantation genetic diagnosis involves removing __________________ from a two-day old embryo after in vitro fertilization.
- A.
Amniotic fluid
- B.
The chorion
- C.
A defective chromosome
- D.
A defective gene
- E.
One cell
Correct Answer
E. One cell -
- 50.
Which type of genetic screening has become mandatory in all states for a subset of preventable and/or treatable genetic diseases?
- A.
Adult screening
- B.
Pre-implantation genetic diagnosis
- C.
Fetal/embryonic screening
- D.
Neonatal screening
Correct Answer
D. Neonatal screening -
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Mar 21, 2022Quiz Edited by
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May 02, 2012Quiz Created by
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