Download Biology 303 EXAM II 3/14/00 NAME

Biology 303 EXAM II 10/27/08
NAME________________________________
-----------------------------------------------------------------------------------------------------------------This exam consists of 40 multiple choice questions worth 2.5 points each. On
the separate answer sheet, please fill-in the single best choice for each question.
Please bubble-in your name on the answer sheet, leaving a space between first
and last names. Also, bubble-in the last 5 digits of your social security number
under “ID NUMBER” beginning in the left-most column. Good luck!
-----------------------------------------------------------------------------------------------------------------1.
In humans, the genetic basis for determining the sex “male” is accomplished by
A. an autosome
B. the number of X chromosomes
C. a balance between the number of X chromosomes and the number of haploid
sets of autosomes
D. none of the above
2.
A Barr body is
A. a gene on the X chromosome that is responsible for female development.
B. a patch of cells that has a phenotype different from surrounding cells because
of variable X inactivation.
C. an inactivated X chromosome, visible in the nucleus of a cell from a female
mammal.
D. someone who hangs out in five-points after class.
3.
What is the expected number of Barr bodies in an individual with the karyotype
XXXYY?
A. 0
B. 1
C. 2
D. 3
4.
The genotype XXYY produces a
A. female in Drosophila and a female in humans.
B. female in Drosophila and a male in humans.
C. male in Drosophila and a male in humans.
D. male in Drosophila and a female in humans.
5.
A somatic cell from a triploid person contains a total of how many
chromosomes?
A. 69
B. 138
C. 3
D. 47
6.
A balanced translocation
A. leads to the condition of “semisterility” even in the absence of any crossingover.
B. leads to semisterility only if a crossover occurs between the translocated
chromosomes during meiosis.
C. greatly increases the chances of nondisjunction.
D. has no consequence whatsoever since the genetic material is still balanced.
7.
Familial Down syndrome is caused by:
A. position effect.
B. nondisjunction.
C. genomic imprinting.
D. a type of translocation.
8.
A human female with Turner syndrome expresses the X-linked trait hemophilia
as did her father. Which of her parents underwent nondisjunction during meiosis, giving
rise to the gamete responsible for the syndrome?
A. her mother
B. her father
C. both parents
D. neither parent
9.
A fertilized egg inherited one X and one Y chromosome, but in a subsequent
round of cell division in the embryo a cell is produced that has lost the Y chromosome.
Which of the following is a possibility regarding the individual that develops from this
embryo?
A. the person may be a hermaphrodite
B. the person has mosaic Turner Syndrome
C. the person may actually have a very mild phenotype.
D. any of the above may be true.
10.
Dosage compensation
A. must occur when one sex has more copies of a gene or genes than the other
sex.
B. is a problem in species that have more autosomes than sex chromosomes.
C. cannot be directed by enhancing X-chromosome activity.
D. works in the same way in all animals.
11.
When it comes to X chromosome inactivation in a human cell, transcription of the
XIST gene is enhanced on
A. the inactivated X chromosome
B. the active X chromosome
C. both X chromosomes
D. the Y chromosome
12.
The observation that a fly heterozygous for the double Bar allele and the wildtype allele (genotype = BD/B+) has a different phenotype than a fly that is homozygous
for the Bar eye allele (genotype = B/B) serves as an example of the phenomenon
known as
A. chaos
B. inversions
C. incomplete penetrance
D. position effect
13.
The genetic material of most living things is
A. deoxyribonucleic acid.
B. ribonucleic acid.
C. RNA.
D. polysaccharide.
14.
Griffith's classic experiments with mice and Diplococcus pneumoniae
A. demonstrated that DNA is the genetic material.
B. demonstrated that the genetic material is not protein.
C. led to the discovery of the "transforming principle."
D. all of the above.
15.
Which of the following clusters of terms accurately describes DNA as it is
generally viewed to exist in prokaryotes and eukaryotes?
A. Double-stranded, parallel, (A+T)/(C+G)= variable, (A+G)/(C+T)=1.0
B. Single-stranded, antiparallel, (A+T)/(C+G)=1.0, (A+G)/C+T)=1.0
C. Double-stranded, antiparallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0
D. Double-stranded, antiparallel, (A+T)/C+G)=1.0, (A+G)/(C+T)=variable
16.
Which of the following is a purine?
A. adenine.
B. cytosine.
C. thymine.
D. alanine.
17.
Which one of the following is part of the Watson-Crick model for DNA?
A. DNA is triple-stranded.
B. the DNA helix is left-handed.
C. DNA consists of two strands of deoxynucleotides with the same polarity.
D. the number of purines equals the number of pyrimidines.
18
In DNA, each nitrogenous base is directly covalently attached to
A. a phosphate group.
B. a deoxyribose.
C. a ribose.
D. none of the above.
19.
The two strands of a DNA helix are held together (to each other) by
1. covalent bonds.
2. hydrogen bonds.
3. phosphodiester bonds.
4. glycosidic bonds.
20.
If one strand of a short DNA fragment has the sequence 5'-AGAGAGAG-3' then
the other strand of DNA has the sequence
A. 5'-TCTCTCTC-3'
B. 5-’GAGAGAGA-3’
C. 3'-AGAGAGAG-3'
D. 3'-TCTCTCTC-5'
21.
Reassociation kinetics experiments reveal that
A. DNA is very easy to melt.
B. DNA cannot be reannealed once melted.
C. the kinetics of reassociation says nothing about genomic structure.
D. eukaryotic genomes contain many repeated sequences.
22.
In the Meselson-Stahl experiment, what was the density distribution of the
isolated DNA molecules two generations after shifting bacteria from "heavy" to "light"
growth medium?
A. 100% of the molecules were of heavy density.
B. 50% were of heavy density, 50% were intermediate density.
C. 100% were of intermediate density.
D. 50% were of light density, 50% were intermediate density.
23.
The enzyme that appears to be inappropriately expressed in cancer cells and
helps to maintain the very ends of eukaryotic chromosomes is called
A. topoisomerase.
B. ligase.
C. telomerase.
D. terminase.
24.
The fact that there is a problem maintaining the very ends of eukaryotic
chromosomes during replication has to do with
A. the fact that eukaryotic chromosomes are linear.
B. the inability of DNA polymerases to initiate synthesis without a primer.
C. the restriction that DNA synthesis must occur in a 5' to 3' direction.
D. all of the above.
25.
During replication of DNA, strand elongation proceeds
A. in a 5' to 3' direction on the leading strand, but in a 3' to 5' direction on the
lagging strand.
B. in a 3' to 5' direction on the leading strand, but in a 5' to 3' direction on the
lagging strand.
C. in a 5' to 3' direction on both the leading and lagging strands.
D. in a 3' to 5' direction on both the leading and lagging strands.
26.
Which of the following is NOT an example of middle repetitive DNA?
A. SINES
B. LINES
C. satellite sequences
D. VNTRs
27.
Satellite sequences are usually found in
A. the centromeric region.
B. the coding region of genes.
C. the introns of genes.
D. bacterial genomes.
28.
With regard to eukaryotic chromatin, when one visualizes "beads-on-a-string"
each bead is actually a
A. gene.
B. histone.
C. nucleosome core particle.
D. 30 nanometer fiber.
29.
In eukaryotic chromatin, the next highest level of organization above "beads-ona-string" is
A. the solenoid, or 30 nm fiber
B. chromatin loops.
C. the nuclear matrix.
D. the metaphase chromosome.
30.
If someone came up to you and said “I have enough DNA in my body to stretch
back and forth between the Earth and Sun 1000 times!,” as a student of BIOL 303 you
would think
A.
B.
C.
D.
this guy’s a freak.
nobody could possibly have that much DNA!
this guy’s a liar.
this guy is correct
31.
In Luria and Delbruck's classic experiments to distinguish between
"spontaneous" versus "adaptive" mutation,
A. evidence for spontaneous mutation was produced.
B. evidence for adaptive mutation was produced.
C. evidence that DNA is the genetic material was produced.
D. all of the above.
32
A mutation which results in multiple amino acid changes in proteins is likely to
be a:
A. frameshift.
B. base substitution.
C. transversion.
D. depurination.
33.
Which statement is true?
A. DNA damage and mutation are the same thing.
B. DNA damage is defined as a permanent change in the nucleotide sequence.
C. although DNA damage doesn't necessarily lead to mutation, when a mutation
does occur it is always caused by DNA damage.
D. DNA damage can lead to mutation, but doesn't always.
34.
An organism that is deficient in the biosynthesis of a particular important nutrient
is called a(n)
A. prototroph.
B. auxotroph.
C. nutritroph.
D. dead organism.
35.
A tautomeric shift
A. is typically induced by high-energy radiation.
B. occurs only in prokaryotes.
C. can lead to a transversion mutation.
D. can produce a transition mutation.
36.
In mammals, DNA double-strand breaks may be repaired accurately by
A. mismatch repair.
B. base excision repair.
C. nucleotide excision repair.
D. homologous recombination
37.
“Nonhomologous end-joining” is
A. a way of repairing a DNA double-strand break that usually produces a
deletion.
B. very similar to base excision repair
C. defective in Xeroderma Pigmentosum
D. a way of repairing AP sites.
38
The simplest transposable elements in bacteria are known as
A. transposons.
B. insertion sequences.
C. Ds elements.
D. Ty elements.
39. Two major classifications of transposable elements in the human genome are
A.
B.
C.
D.
SINEs and satellites
VNTRs and STRs
Copia and P elements
DNA transposons and retrotransposons
40.
The presence of transposons in a genome can have an impact on gene
expression by
A. inducing gene silencing
B. disrupting a gene by inserting into a coding region
C. creating the potential for aberrant chromosomal rearrangements
D. all of the above
That
wasn’t so
scary!
ANSWER KEY
Q: 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14,15,16,17,18,19,20,
A: D, C, C, B, A, A, D, A, D, A, A, D, A, C, C, A, D, B, B, D,
Q: 21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40
A: D, D, C, D, C, C, A, C, A, D, A, A, D, B, D, D, A, B, D, D