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Williams Bio 93 Midterm 3 Fall 2014
Exam Key
1. What is the outermost layer of the Ebola virus?
A. Capsid
B. Capsomere
C. Lipid membrane
D. RNA molecule
2. The antiviral drug ribavirin has not seen widespread use because of severe side effects. It acts like a
guanosine and blocks cell functions that require GTP and guanine nucleotides. Which of the following will
NOT be affected?
A. Translation
B. Binding of transcription factors
C. RNA synthesis
D. G-protein receptor function
3. One of the possible treatments for Ebola is an antiviral drug called
favipiravir. The researchers think that the favipiravir may be acting like a
“fake” nucleotide that blocks replication if there are few other nucleotides
of the same type. To test this, they added the drug and high quantities of
other nucleotides. Which type of nucleotide is the drug normally acting as?
A. A purine
B. A pyrimidine
Adding lots of adenine and guanine restored replication, and they are purines. So the drug is a fake purine.
4. An antiviral drug that stops a filovirus but does not harm the infected cell would best act on:
A. Ribosomes encased in the viral capsid
B. Polymerase that transcribes RNA
C. Viral DNA that contains uracil
D. Proton pumps in the cellular lysosomes
Cells to not transcribe RNA, only viruses do this.
5. According to the figure, the effect of molecule p21 is to:
A. Break phosphodiester bonds
B. Increase rate of mitosis
C. Reduce DNA synthesis
The following figure is from a study examining the effects of an anti-cancer drug on a group of dividing human
cancer cells. The next three questions relate to this.
6. The X axis indicates the amount of fluorescently labeled DNA within each cell, while the Y axis counts the
number of cells with this fluorescence. This graph is an example of a:
A. Histogram
B. Line graph
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Williams Bio 93 Midterm 3 Fall 2014
C. Correlation
7. The cells found under the peak at G1:
A. Contain 46 sister chromatids
B. Contain 23 homologous pairs of chromosomes
C. Contain two nuclei
D. Contain fully condensed chromosomes
No synthesis yet, and no mitosis
8. The (b) figure is the control, the (d) figure is the result of application of the cancer
drug. Which of the following is most likely true?
A. The drug blocks DNA synthesis
B. The drug slows DNA synthesis
C. The drug prevents mitosis
D. The drug reduces the number of cells
Lots of cells are past S phase, so DNA synthesis is going fine. Cell number is not directly measured. Since there is
evidence of polyploidy, mitosis is not following division.
9. Which of the following is true in metaphase of meiosis I but NOT in metaphase of mitosis?
A. Microtubules from both spindle poles attach to each centromere
B. Sister chromatids are recombinant
C. Chromosomes line up randomly at the metaphase plate
10. Which of the following would signal nondisjunction in meiosis 1 rather than meiosis 2?
A. All normal gametes
B. Half normal gametes
C. No normal gametes
D. Normal gametes but abnormal development
E. Normal gametes but a reduced number of gametes
If the division was meiosis 2, half the daughter cells would be normal.
11. A man has a DNA replication error causing a deletion of DNA in a reproductive cell about to enter meiosis
that will produce sperm cells. A woman’s skin cell acquires a substitution mutation during replication prior to
mitosis due to the sun’s radiation. If these two individuals mate, which of the mutations might be passed
onto their children?
A. Neither mutation has a chance to be passed on
B. The man’s deletion mutation has a chance to be passed on
C. The woman’s substitution mutation has a chance to be passed on
D. Both mutations have a chance to be passed on
Somatic cell mutations are not passed on.
12. The process of PCR is used to synthesize new DNA in the lab. The doublestrand of DNA is separated (“melted”) by adding heat to the DNA solution. It
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Williams Bio 93 Midterm 3 Fall 2014
has become clear that DNA with different percentages of base pairs vary in how easily they melt, as shown in
this figure. Which is most likely true?
A. G/C rich strands form looser bonds than A/T rich strands
B. PCR is normally performed at human body temperature (37 oC)
C. The process of PCR breaks phosphodiester bonds
D. The amount of hydrogen bonding determines the melting point of DNA
13. There is a eukaryotic cell whose nuclear envelope only allows molecules to enter the nucleus. Which of the
following processes will not be able to take place in this cell?
A. tRNAs binding to codons
B. DNA replication
C. RNA being made from DNA
D. tRNAs binding to amino acids
All others happen to molecules in the nucleus or to molecules in the cytoplasm that weren’t in the nucleus.
14. A nonsense mutation in a gene:
A. Alters the reading frame of the mRNA
B. Changes an amino acid in the encoded protein
C. Has no effect on the amino acid sequence of the encoded protein
D. Introduces a stop codon into the mRNA
E. Prevents introns from being expressed
Nobel Prize winner David Baltimore was studying an RNA virus that was able to cause chicken cells to divide and
form tumors. He suspected the presence of a reverse transcriptase enzyme that was present in the virus and
transcribed the viral RNA to cellular DNA. The next two questions relate to this.
15. In his first experiment, he removed the capsids from several RNA viruses and added the virus contents to two
vials (vial X and vial Y) with free cytosine, adenine, guanine, uracil and radioactive thymine. To vial X he
added an RNase (breaks down RNA). To vial Y he added a DNase (breaks down DNA). What was the likely
result of this experiment?
A. Radioactive nucleic acid formed, destroyed by RNase but not DNase
B. Radioactive nucleic acid formed, destroyed by DNase but not RNase
C. Non-radioactive nucleic acid formed, destroyed by RNase but not DNase
D. Non-radioactive nucleic acid formed, destroyed by DNase but not RNase
16. In his second experiment, Baltimore placed the RNA virus in a test tube (vial Z) and added trypsin, which
breaks down all proteins in the virus. He then added all five nucleotides (including radioactive thymine) to the
viral material that remained. Why did no nucleic acids form?
A. The capsid was destroyed
B. The RNA was destroyed
C. The ribosomes were destroyed
D. The reverse transcriptase was destroyed
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Williams Bio 93 Midterm 3 Fall 2014
17.
A template DNA contains this sequence:
AUG
3’ – GATACATGAGGCGAT – 5’
1 2 3
Which location is the complement to where translation of the associated mRNA will begin? It may help to
write the mRNA above the DNA.
A. The G at site 1
B. The T at site 2
C. The A at site 3
18. In current scientific literature, cis-regulatory elements are gene regulation elements found on the same
chromosome as the target gene, while trans-acting factors are proteins that may be coded by genes on other
chromosomes. Which of these is a trans-acting factor?
A. Enhancer
B. TATA box
C. Promoter
D. Transcription factor
19. The stickleback fish carries a gene called Pitx1 that codes for a protein that triggers the development of the
pituitary gland, the bones of the jaw, and spines in the pelvic fins in baby fish. Fish that live in small lakes tend
to lose their pelvic spines over thousands of generations, but still maintain a functional jaw and pituitary
gland. Which of the following could be the reason for the loss of the pelvic spines?
A. Deletion of the Pitx1 gene or promoter
B. Mutation in the pelvic Pitx1 enhancer or activator
C. Deletion of the pituitary Pitx1 gene in the cells in the pelvic region
D. Translocation in the chromosome with the Pitx1 gene
20. Spineless fish from a small lake were mixed with spined fish from larger lakes. Their offspring (baby fish) are
called hybrids because they have one spined and one spineless parent. The amount of Pitx1 activator and
mRNA in the jaw tissue and pelvic tissue was measured. The following results were obtained:
In the Jaw
In the Pelvis
Amt of Pitx1
Amt of Pitx1
Amt of Pitx1
Amt of Pitx1
activator
mRNA
activator
mRNA
Spineless fish
normal
normal
low
low
Hybrid fish
normal
normal
normal
low
Which hypothesis do these data support?
A. A mutation in the pelvic enhancer gene
B. A mutation in the pelvic activator gene
C. A mutation in the Pitx1 gene
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