Lecture 1-9 for 100 Describe a Structural Difference between
DNA and RNA Answer
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Lecture 1-9 100 Answer: What is C) DNA nucleotides contain a
different sugar than RNA nucleotides. DNA has deoxyribose, and RNA
has the ribose
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Lecture 1-9 for 200 One liter of a solution of pH 2 has how
many more hydrogen ions (H ) than 1 L of a solution of pH 6?
Answer
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Lecture 1-9 200 Answer What is 10 4 = 10,000 times more
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Lecture 1-9 for 300 Each organism is uniquely adapted to its
environment even at the molecular level. Human fats are about 3%
polyunsaturated while that of cod, a fish living in cold ocean
waters, is approximately 50% polyunsaturated. Knowing this, why can
the cod survive colder temperatures than a human? A) Unsaturated
fats act as hormones activating genes for proteins that act as
insulators. B) Unsaturated fats dissolve in the aqueous portions of
cells acting like antifreeze and slow the formation of ice
crystals. C) Unsaturated fats remain fluid at lower temperatures
than saturated fats due to their shape. D) Unsaturated fats can be
assembled into thicker cell membranes providing additional
insulation against the cold Answer
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Lecture 1-9 for 300 Answer What is C) Unsaturated fats remain
fluid at lower temperatures than saturated fats due to their
shape
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Lecture 1-9 for 400 There are 20 different amino acids. What
makes one amino acid different from another? A) different carboxyl
groups attached to an alpha carbon B) different amino groups
attached to an alpha carbon C) different alpha carbons D) different
side chains (R groups) attached to an alpha carbon Answer
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Lecture 1-9 for 400 Answer What is D) different side chains (R
groups) attached to an alpha carbon
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Lecture 1-9 for 500 What are the 3 phases of cell signaling?
Answer
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Lecture 1-9 for 500 Answer 1)Reception 2)Transduction (Relay)
3)Response
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Lecture 1-9 for 600 Answer What are the 4 kinds of signal
types?
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Lecture 1-9 for 600 Answer 1)Paracrine 2)Endocrine 3)Cell
Surface 4)Intracellular
Slide 21
Lecture 10-18 for 100 Answer What is the difference between an
exergonic and endergonic reaction?
Slide 22
Lecture 10-18 for 100 Answer Exergonic releases energy;
products more stable than reaction Endergonic absorbs energy;
nonspontaneous products less stable
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Lecture 10-18 for 200 Answer
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Lecture 10-18 for 200 Answer What is E) mitochrondrion
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Lecture 10-18 for 300 Answer What role do NAPH and FADH2 play
in cellular respiration?
Slide 26
Lecture 10-18 for 300 Answer Electron carriers
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Lecture 10-18 for 400 Answer What is the difference between a
oncogene and a protooncogene?
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Lecture 10-18 for 400 Answer Proto-oncogene: normal gene
involved in cell cyle regulation Oncogene: mutate
protooncogene
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Lecture 10-18 for 500 Answer
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Lecture 10-18 for 500 Answer What is C) glucose is oxidized and
oxygen is reduced Glucose loses electron and Oxygen gains electron
(OIL RIG) Remember, this is not the same as oxidizing agent and
reducing agent! Oxidizing agent gains e- and gets reduced Reducing
agent loses e- and gets oxidized
Slide 31
Lecture 10-18 for 600 Answer What phases of cell cycle do law
of independent assortment and law of segregation affect?
Slide 32
Lecture 10-18 for 600 Answer Independent Assortment: Metaphase
I (depends on genes being on separate chromosome) Segregation
Anaphase I depends on alleles segregating into different games
(each sister chromatid has copy of allele)
Slide 33
Lecture 19-22 for 100 Which would you expect of a eukaryotic
cell lacking telomerase? A) a high probability of becoming
cancerous B) production of Okazaki fragments C) inability to repair
thymine dimers D) a reduction in chromosome length E) high
sensitivity to sunlight Answer
Slide 34
Lecture 19-22 for 100 Answer What is D) a reduction in
chromosome length Telomerase is an enzyme that adds DNA sequence
repeats to the 3 end of DNA strands in the telomere regions
Slide 35
Lecture 19-22 for 200 Which of the following types of mutation,
resulting in an error in the mRNA just after the AUG start of
translation, is likely to have the most serious effect on the
polypeptide product? A) a deletion of a codon B) a deletion of 2
nucleotides C) a substitution of the third nucleotide in an ACC
codon D) a substitution of the first nucleotide of a GGG codon E)
an insertion of a codon Answer
Slide 36
Lecture 19-22 for 200 Answer What is B) a deletion of 2
nucleotides
Slide 37
Lecture 19-22 for 300 Polytene chromosomes of Drosophila
salivary glands each consist of multiple identical DNA strands that
are aligned in parallel arrays. How could these arise? A) meiosis
followed by mitosis B) fertilization by multiple sperm C)
replication followed by mitosis D) replication without separation
E) special association with histone proteins Answer
Slide 38
Lecture 19-22 for 300 Answer What is D) replication without
separation
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Lecture 19-22 for 400 Each of the following options is a
modification of the sentence THECATATETHERAT. Which of the
following is analogous to a frameshift mutation? A) THERATATETHECAT
B) THETACATETHERAT C) THECATARETHERAT D) THECATATTHERAT E)
EHHCATATETHERAT Answer
Slide 40
Lecture 19-22 for 400 Answer What is D) THECATATTHERAT There is
a single deletion of E, which would cause a shift in the reading
frame
Slide 41
Lecture 19-22 for 500 When DNA is compacted by histones into 10
nm and 30 nm fibers, the DNA is unable to interact with proteins
required for gene expression. Therefore, to allow for these
proteins to act, the chromatin must constantly alter its structure.
Which processes contribute to this dynamic activity? A) DNA
supercoiling at or around H1 B) hydrolysis of DNA molecules where
they are wrapped around the nucleosome core C) accessibility of
heterochromatin to phosphorylating enzymes D) nucleotide excision
and reconstruction E) methylation and phosphorylation of histone
tails Answer
Slide 42
Lecture 19-22 for 500 Answer What is E) methylation and
phosphorylation of histone tails Methylation and phosphorylation to
histone tails alter chromatin structure
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Lecture 19-22 for 600 A mutant bacterial cell has a defective
aminoacyl synthetase that attaches a lysine to tRNAs with the
anticodon AAA instead of a phenylalanine. The consequence of this
will be that A) none of the proteins in the cell will contain
phenylalanine. B) proteins in the cell will include lysine instead
of phenylalanine at amino acid positions specified by the codon
UUU. C) the cell will compensate for the defect by attaching
phenylalanine to tRNAs with lysine-specifying anticodons. D) the
ribosome will skip a codon every time a UUU is encountered. E) None
of the above will occur; the cell will recognize the error and
destroy the tRNA. Answer
Slide 45
Lecture 19-22 for 600 Answer What is B) proteins in the cell
will include lysine instead of phenylalanine at amino acid
positions specified by the codon UUU
Slide 46
Lecture 23-26 for 100 The bicoid gene product is normally
localized to the anterior end of the embryo. If large amounts of
the product were injected into the posterior end as well, which of
the following would occur? A) The embryo would grow to an unusually
large size. B) The embryo would grow extra wings and legs. C) The
embryo would probably show no anterior development and die. D)
Anterior structures would form in both sides of the embryo. E) The
embryo would develop normally. Answer
Slide 47
Lecture 23-26 100 Answer What is D) Anterior structures would
form in both sides of the embryo
Slide 48
Lecture 23-26 for 200 Answer During the early part of the
cleavage stage in frog development, the rapidly developing cells A)
skip the mitosis phase of the cell cycle. B) skip the S phase of
the cell cycle. C) skip the G1 and G2 phases of the cell cycle. D)
rapidly increase the volume and mass of the embryo. E) skip the
cytokinesis phase of the cell cycle.
Slide 49
Lecture 23-26 for 200 Answer What is C) skip the G1 and G2
phases of the cell cycle Frog cleavage cycles consists of S phase
and M phase but lack G1 and G2 phases
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Lecture 23-26 for 300 If gastrulation was blocked by an
environmental toxin, then A) embryonic germ layers would not form.
B) cleavage would not occur in the zygote. C) fertilization would
be blocked. D) the blastula would not be formed. E) the blastopore
would form above the gray crescent in the animal pole. Answer
Slide 52
Lecture 23-26 for 300 Answer What is A ) embryonic germ layers
would not form
Slide 53
Lecture 23-26 for 400 If a Drosophila female has a homozygous
mutation for a maternal effect gene, A) she will not develop past
the early embryonic stage. B) only her male offspring will show the
mutant phenotype. C) her offspring will show the mutant phenotype
only if they are also homozygous for the mutation. D) only her
female offspring will show the mutant phenotype. E) all of her
offspring will show the mutant phenotype, regardless of their
genotype. Answer
Slide 54
Lecture 23-26 for 400 Answer What is E) all of her offspring
will show the mutant phenotype, regardless of their genotype. With
maternal effect gene, an organism shows the phenotype expected from
the genotype of the mother, irrespective of its own genotype
Slide 55
Lecture 23-26 for 500 This segment of DNA 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)E) B)D)
Answer
Slide 56
Lecture 23-26 for 500 Answer What is B) DNA is negatively
charged due to the phosphate ions present in the ribose-phosphate
backbone. It moves towards the positive pole during
electrophoresis. The shorter the DNA fragment, the faster it moves
towards the positive pole.
Slide 57
Lecture 23-26 for 600 For a neuron with an initial membrane
potential at - 70 mV, an increase in the movement of potassium ions
out of that neuron's cytoplasm would result in A) depolarization of
the neuron. B) hyperpolarization of the neuron. C) the replacement
of potassium ions with sodium ions. D) the replacement of potassium
ions with calcium ions. E) the neuron switching on its
sodium-potassium pump to restore the initial conditions.
Answer
Slide 58
Lecture 23-26 for 600 Answer What is B) hyperpolarization of
the neuron.