Download Williams Bio 93 Final Exam Fall 2014 Answer Key 1

Williams Bio 93 Final Exam Fall 2014 Answer Key
Use the adjacent image (chloroplast above, mitochondrion below) to answer the next questions.
1. What is the oxidized molecule at position L?
A. Carbon dioxide
B. Sugar
C. Water
D. Oxygen
Carbon dioxide is the only oxidized product used by the chloroplast
2. What molecule is equal to P in the diagram?
A. M
B. N
C. O
D. Q
The only reactant used in respiration that also appears in the cycle is oxygen, given off by chloroplasts.
3. Which represents the product of an endergonic reaction?
A. O
B. L
Sugars have higher free energy than carbon dioxide and water.
This image indicates chromosomes during cell division:
4. Which is correct?
A. J and K are homologous chromosomes
B. L and M are homologous chromosomes
5. Which of the following could take place in either mitosis or meiosis?
A. Image G
B. Image H
Image G can be Meiosis 2 or Mitosis metaphase
6. Some researchers have hypothesized that a carrier for cystic fibrosis (CF) may be protected from cholera.
Why is this a logical hypothesis?
A. CF carriers have a reduced number of CFTR proteins
B. Cholera is caused by the same mutation as CF
C. Both cholera and CF depend upon the presence of gut bacteria
D. Both cholera and CF require chloride to enter the cell
Fewer CFTR proteins means the cell cannot lose as much chloride or water
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Williams Bio 93 Final Exam Fall 2014 Answer Key
7. G6PD Deficiency in humans is an X-linked recessive disease. If a homozygous dominant female (XGXG) has
children with an affected male (XgY), then what is the probability of the children being carriers?
A. All children will be carriers
B. None of the children will be carriers
C. Only the daughters will be carriers
D. Only the sons will be carriers
E. Only half of the daughters will be carriers
The disease will be given by the man to all his daughters and none of his sons.
8. In the cross PpRRYy X PpRrYY, what fraction of the offspring will have white flowers and round, yellow
seeds? (P = dominant allele for purple; R = dominant allele for round; Y = dominant allele for yellow)
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
1/4 x 1 x 1 = 1/4
9. The traits of red hair and increased skin cancer are associated with a mutation of the Mc1r gene. This is an
example of:
A. Pleiotropy
B. Polygenic inheritance
C. Codominance
10. There are up to seven variants of Mc1r that are associated with red hair, a recessive trait. If dark-haired
parents have two red-haired children, what is the maximum number of red-haired alleles present in the
family?
A. One
B. Two
C. Four
D. Eight
Dark-haired parents can have one recessive allele each, and they can be different alleles. That’s two total.
11. Mc1r codes for a G protein receptor in melanocytes in all mammals. In the absence of the receptor, yellowred phenomelanin pigment is produced. If the receptor is present and the protein beta defensin is present in
the blood, the beta defensin binds to Mc1r and activates the production of eumelanin, a black pigment.
Which is LEAST likely true?
A. The beta defensin gene and Mc1r gene have an epistatic relationship
B. Black dogs have more cyclic AMP present in their melanocytes.
C. Phenomelanin and eumelanin are codominant
D. Yellow dogs can have either a mutant Mc1r OR a mutant beta defensin
They are not both produced simultaneously.
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Williams Bio 93 Final Exam Fall 2014 Answer Key
The disease typhoid is caused by bacteria called S. typhi. Like cholera, the disease is spread by fecal material in
the water. In typhoid, the bacteria enter the cells of the gut. There is some evidence that people who are
resistant to typhoid often have one mutant CFTR allele. The next six questions use this information.
12. What characterizes people with one mutant ΔF508 CFTR allele (the most common mutation)?
A. Have only one Chromosome 7, the chromosome containing the CFTR gene
B. Have half the normal amount of functional CFTR
C. Half their children will have CF
13. The ΔF508 mutation for CFTR is characterized by the loss of three base pairs. This is what type of mutation?
A. Frameshift
B. Missense
C. Nonsense
D. Deletion
14. S. typhi bacteria enter a gut cell via endocytosis. What is the major component of the endosome membrane
surrounding the bacteria?
A. Phospholipid
B. Protein
C. carbohydrate
Researchers investigated this link between CF carriers and resistance to typhoid with the following experiment.
They created three groups of cells, and exposed each group to S. typhi bacteria with the result shown.
15. What does this figure indicate?
A. Wild-type CFTR protects the cells from the typhoid bacteria
B. Typhoid bacteria are unable to bind to cells at body temperature (37oC)
C. Cells with functional CFTR channels show more internalization of S.
typhi.
Lots of bacteria get into cells at 37 oC if they have wild-type CFTR
16. Why was CFTR cDNA used instead of the CFTR gene?
A. The cDNA is more accurate
B. The cDNA has no introns
C. The cDNA contains the necessary promotors
Believing that the typhoid bacteria are binding to the receptor itself, the researchers conducted a new
experiment. They mixed S. typhi bacteria with colon cancer cells, which have large quantities of wild-type CFTR.
To this mixture they added increasing amounts of plasma membrane fragments removed from three different
cell types:
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Williams Bio 93 Final Exam Fall 2014 Answer Key
mouse cells with no CFTR (light grey),
mouse cells with ΔF508 CFTR (dark grey),
or mouse cells with wild-type CFTR (black).
The white bar is a control.
17. What do these results indicate?
A. In the presence of wild-type CFTR on membrane fragments, S. typhiI are less
likely to bind to cell CFTR
B. The S. typhi bacteria are unable to enter cells when plasma membrane fragments are present
C. S. typhi bacteria bind to membranes that do not contain any CFTR.
The bacteria are binding to the wild-type CFTR on all the fragments, so fewer are entering the actual cells
18. A woman has a daughter who is blood type O and a son who is blood type A. Which of the following are
possible genotypes for the mother?
A. IAIA or IAi
B. IAIB or IAi
C. ii or IBi
Mom needs to give an i to both children.
19. A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the
mRNA codon is
A. TTT
B. UUA
C. UUU
D. AAA
The anticodon matches the coding sequence, but with uracils.
“Grasshopper” mice eat scorpions, and unlike regular house mice, show little pain if stung with scorpion venom.
Here is the representation of a sensory neuron that responds to pain at the skin and synapses onto other neurons
in the spinal cord and brain. The anatomy is somewhat different than the “standard” neuron.
20. When a mouse is stung by a scorpion, which direction will the action
potential move?
A. Toward the pain receptors
B. Toward the synaptic terminals
21. An important membrane channel in pain sensing is the voltage-gated sodium channel called Nav1.8. This
channel is a:
A. Passive transporter synthesized on bound ribosomes
B. Passive transporter synthesized on free ribosomes
C. Active transporter synthesized on bound ribosomes
D. Active transporter synthesized on free ribosomes
Sodium moves down its gradient, protein is synthesized in the membrane of the rER
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Williams Bio 93 Final Exam Fall 2014 Answer Key
22. The neurotransmitter released at the synaptic terminals of a sensory neuron is the amino acid glutamate.
What is the R group?
A. A
B. B
C. C
Find the alpha carbon between the amino and carboxyl groups.
23. The enzymes needed for manufacture of glutamate are made in the mitochondria. How are mitochondria
transported to the synaptic terminals?
A. Via kinesin on microtubules
B. Active transport using sodium potassium pumps
C. Via endocytosis and lysosomes
24. The signalling of one neuron to a second neuron is an example of:
A. Intracellular signaling
B. Cell-cell recognition
C. Local signaling
D. Hormone signaling
The following is a representation of the voltage-gated sodium channel “Nav1.8” found in both house mice and
grasshopper mice.
25. Which label points to amino acids that are exposed to a higher concentration of sodium ions when the
channel is closed?
A. X
B. Y
C. Z
Sodium concentration is higher outside the cell.
26. What is the bond indicated by label X?
A. Peptide bond formed by dehydration
B. Phosphodiester bond formed by hydrolysis
C. Ester bond formed by dehydration
D. Glycosidic bond formed by hydrolysis
27. Which amino acids are more likely to be nonpolar?
A. Amino acids near X
B. Amino acids near Y
These amino acids are inside the lipid bilayer.
The genes for Nav1.8 for house mice (Mus musculus) and grasshopper mice
(Onychomys torridus) were sequenced, and found to both contain four membranespanning domains: OI – OIV for grasshopper mice and MI – MIV for house mice. But the
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Williams Bio 93 Final Exam Fall 2014 Answer Key
gene sequence is different in the different mouse species.
28. These two sequences of genes are:
A. On homologous chromosomes
B. Alleles of the same gene
C. Different genes
D. On different chromosomes
In neurons that transmit pain signals, a painful stimulus normally triggers the opening of the Nav1.8 sodium
channel. Researchers measured the effect of scorpion venom on different versions of the Nav1.8. In (C) they
measure sodium flow (current) in cells with the two wild-type Nav1.8 channels (O = grasshopper mice, M = house
mice), in (E) they measure the effect of changing one segment of the gene sequence in the house mice Nav1.8.
29. Figure C indicates that at high concentrations of scorpion venom, the neurons of grasshopper mice are:
A. More depolarized than house mice
B. Less depolarized than house mice
Less current passes through channel, so less depolarization occurs.
30. Figure D indicates that modifying one section of the house mouse Nav1.8 gene causes:
A. Increased flow of sodium at high venom concentration
B. Decreased flow of sodium at low venom concentration
C. The house mouse Nav1.8 to act like the grasshopper mouse Nav1.8.
Decreased flow of sodium at high venom concentration
31. An antiviral drug being tested on Ebola acts like a purine during the copying of negative-strand RNA. It binds
to the growing nucleic acid but halts further translation. This binding-and-blocking technique is also used in:
A. PCR
B. cDNA
C. Gel electrophoresis
D. Sanger DNA sequencing
32. A change commonly referred to as a mutation occurs to a DNA molecule in an organism's liver cell before the
organism reproduces. When the organism reproduces, how many of its offpsring will have the mutation?
A. It depends on how many offspring the organism has
B. All of the organism's offspring will have the mutation
C. Some of the organism's offspring will have the mutation
D. None of the organism's offspring will have the mutation
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Williams Bio 93 Final Exam Fall 2014 Answer Key
Somatic cells do not pass on mutations, only sex cells do.
33. The DNA molecules in eye cells contain information about which of the following?
A. Neither eye color nor skin color
B. Eye color and skin color
C. Skin color, but not eye color
D. Eye color, but not skin color
All cells still contain all DNA, but only some genes are expressed
The NPC1 protein is a receptor found in endocytosed vesicles that have brought in lipids from outside the cell, as
shown. Researchers have found that blocking NPC1 prevents Ebola infection. This information is used for the next
six questions.
34. Where in the cycle of Ebola infection does this block occur?
A. Binding to the cell membrane (accepted)
B. Entry into the cell
C. Entry into cytoplasm (correct)
D. Transcription of RNA
E. Translation of viral protein
Ebola binds to unknown receptors on the cell membrane and is endocytosed. The virus THEN binds to NPC1
within the endosome vesicle, a receptor normally used only for lipid binding and release. It utilizes the NPC1 to
exit the endosome. But since this wasn’t completely clear in the question, A was accepted as a part of the process
we had already discussed. C is the correct biological answer.
35. Individuals that are homozygous for a mutation of the NPC1 gene have a fatal disease called Niemann-Pick
Disease Type C, which is characterized by large amounts of lipid-filled liposomes filling the cell. If two
individuals who are carriers for the mutation have two children, what is the likelihood both children will have
Niemann-Pick?
A. 1/16
B. 1/8
C. 1/4
D. 1/2
1/4 x 1/4 = 1/16
36. Researchers studying the effects of toxic environmental hydrocarbons (called
TCDDs) on expression of NPC1 found the effect shown. What is the best
description of these findings?
A. Transcription of NPC1 is reduced by increasing levels of hydrocarbons
B. Hydrocarbons cause mutation of the NPC1 gene
C. Hydrocarbons add acetyl groups to histones near the NPC1 gene
Less mRNA produced means less transcription
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Williams Bio 93 Final Exam Fall 2014 Answer Key
37. In order to determine how long the NPC1 mRNA remain in the cell, researchers next blocked new mRNA
production using the drug Actinomycin D (Act-D). Where should Act-D bind in order to block mRNA
production?
A. DNA polymerase
B. Ribosomes
C. General transcription factors
D. Restriction enzyme
Only general transcription factors are associated with mRNA synthesis
38. The researchers found the results shown when comparing the amounts
of mRNA in cells treated with Act-D alone, or cells treated with Act-D
and hydrocarbon TCDD. There was no significant difference between
the treatments. This indicates the environmental hydrocarbons act by:
A. Blocking transcription
B. Increasing splicosomes
C. Increasing miRNA
No evidence of posttranslational reduction compared to control, so blocking is at transcription.
39. Based on these results above, it is possible that people who live near chemical plants that produce
environmental hydrocarbons would have:
A. Increased lipid storage disorders and reduced Ebola infections
B. Reduced lipid storage disorders and increased Ebola infections
40. What is it about nucleic acids that makes copying genetic information straightforward?
A. The binding of bases to one another is specific.
B. The sequence of bases encodes information.
C. Hydrogen bonds are easily broken.
D. The shape of the molecule is determined by the information it contains.
Complementary binding is the basis of both transcription and translation.
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