Biology
Bio 93 Final Exam
Review
Lecture
1-9
Lecture
10-18
Lecture
19-22
Lecture
23-26
100
100
100
100
200
200
200
200
300
300
300
300
400
400
400
400
500
500
500
500
600
600
600
600
Lecture 1-9 for 100
Describe a Structural Difference between DNA
and RNA
Answer
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
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
Lecture 1-9 200 Answer
What is 104 = 10,000 times more
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
Lecture 1-9 for 300 Answer
What is C) Unsaturated fats remain fluid at lower
temperatures than saturated fats due to their shape
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
Lecture 1-9 for 400 Answer
What is D) different side chains (R groups) attached to
an alpha carbon
Lecture 1-9 for 500
What are the 3 phases of cell signaling?
Answer
Lecture 1-9 for 500 Answer
1) Reception
2) Transduction (Relay)
3) Response
Lecture 1-9 for 600
What are the 4 kinds of signal types?
Answer
Lecture 1-9 for 600 Answer
1)
2)
3)
4)
Paracrine
Endocrine
Cell Surface
Intracellular
Lecture 10-18 for 100
What is the difference between an exergonic and endergonic reaction?
Answer
Lecture 10-18 for 100 Answer
Exergonic releases energy; products more stable than reaction
Endergonic absorbs energy; nonspontaneous products less stable
Lecture 10-18 for 200
Answer
Lecture 10-18 for 200 Answer
What is E) mitochrondrion
Lecture 10-18 for 300
What role do NAPH and FADH2 play in cellular respiration?
Answer
Lecture 10-18 for 300 Answer
• Electron carriers
Lecture 10-18 for 400
• What is the difference between a oncogene
and a protooncogene?
Answer
Lecture 10-18 for 400 Answer
Proto-oncogene: normal gene involved in cell
cyle regulation
Oncogene: mutate protooncogene
Lecture 10-18 for 500
Answer
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
Lecture 10-18 for 600
What phases of cell cycle do law of independent assortment and law of segregation affect?
Answer
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)
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
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
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
Lecture 19-22 for 200 Answer
What is B) a deletion of 2 nucleotides
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
Lecture 19-22 for 300 Answer
• What is D) replication without separation
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
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
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
Lecture 19-22 for 500 Answer
What is E) methylation and phosphorylation of
histone tails
Methylation and phosphorylation to histone tails
alter chromatin structure
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
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
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
Lecture 23-26 100 Answer
What is D) Anterior structures would form in both sides
of the embryo
Lecture 23-26 for 200
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.
Answer
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
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
Lecture 23-26 for 300 Answer
What is A) embryonic germ layers would not form
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
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
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)
B)
D)
E)
Answer
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.
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
Lecture 23-26 for 600 Answer
What is B) hyperpolarization of the neuron.