BIOL . 303 EXAM II 10/20/05 VERSION 1 Name________________________
-----------------------------------------------------------------------------------------------------------------This exam consists of 40 multiple choice questions worth 2.5 points each. On
the separate testing form, please fill-in the single best choice for each question. Be
sure to fill-out your student number and name on the answer sheet. Also, write
“version 1” in the space for “college.” Good luck!
-----------------------------------------------------------------------------------------------------------------1.
It is usually easier to identify spontaneous mutations in bacteria than most
eukaryotes because
1. mutations are visible as color changes within a bacterial plaque.
2. mutations can be induced via exposure to the bacteriophage T1.
3. given the haploid condition of most bacteria, mutations are phenotypically
expressed readily in bacterial cells.
4. mutations occur at a much higher frequency in bacteria.
2.
In conjugation, genetic information is transferred
1. from one cell into the culture medium, where it is taken up by another cell.
2. with the help of a viral go-between.
3. in a bidirectional fashion between two cells.
4. from one bacterium to another.
3.
Hfr bacterial cells exhibit the highest levels of gene transfer because
1. they contain two distinct F factors.
2. they induce the production of significantly more sex pili than normal F+ cells.
3. rather than existing as an isolated plasmid, the F factor is incorporated within
the main chromosome where it can transfer many genes at one time.
4. they do not require conjugation for gene transfer.
4.
In cotransformation
1. non-linked genes are passed simultaneously from one cell to another.
2. a bacterial cell receives two adjacent genes on a single piece of DNA from the
medium.
3. two bacterial cells within a culture are transformed by the same genetic
material.
4. it is not uncommon for the entire bacterial chromosome and F factor to be
passed from one cell to another.
5.
The name for the general category of DNA molecules into which double-stranded
circular extrachromosomal DNA elements such as F factors would fall is
1. capsid
2. plasmid
3. plaque
4. partial diploid
6.
In humans, the genetic basis for determining the sex “male” is accomplished by
the presence of
1. the Y chromosome
2. one X chromosome
3. a balance between the number of X chromosomes and the number of haploid
sets of autosomes
4. multiple alleles scattered throughout the autosomes
7.
Klinefelter syndrome in humans, which leads to underdeveloped testes and
sterility, is caused by which chromosomal condition?
1.
2.
3.
4.
47, XXY
47, 21+
45, X
47, XYY
8.
In birds, sex is determined by a ZW chromosome scheme, which is much like the
typical XY scheme seen in humans and many other organisms. However, in birds the
system is reversed. Males are ZZ (similar to XX in humans) and females are ZW
(similar to XY in humans). A lethal recessive allele that causes death of the embryo
occurs on the Z chromosome in pigeons. What would be the sex ratio in the offspring of
a cross between a male heterozygous for the lethal allele and a normal female?
1. 1:1 male to female
2. 3:1 male to female
3. 1:2 male to female
4. 2:1 male to female
9.
Hemophilia is caused by several genetic factors; one, a sex-linked recessive
gene, is the subject of this problem. Assume that a man with hemophilia marries a
normal woman whose father had hemophilia. What is the probability that they will have
a daughter with hemophilia? (Note: in this problem you must include the probability of
having a daughter in your computation of the final probability.)
1. 1/16
2. 1/8
3. 1/4
4. 1/2
10.
What is the expected number of Barr bodies in an individual with the karyotype
XXYYY?
1. 0
2. 1
3. 2
4. 3
11.
When an organism gains or loses one or more chromosomes but not a complete
haploid set, the condition is known as
1. polyploidy
2. aneuploidy
3. triploidy
4. trisomy
12.
The condition known as cri-du-chat syndrome in humans has a genetic
constitution designated as.
1. 45, X
2. 46,5p3. triploidy
4. trisomy
13.
A genomic condition that may be responsible for some forms of fragile-X
syndrome, as well as Huntington disease, involves
1. F plasmids inserted into the FMR-1 gene.
2. expansion of trinucleotide repeats.
3. multiple inversions in the X chromosome.
4. single translocations in the X chromosome.
14.
Trisomy 21, or Down syndrome, occurs when there is a normal diploid
chromosomal complement of 46 chromosomes plus one (extra) chromosome #21.
Such individuals therefore have 47 chromosomes. While there is impaired fertility of
both sexes, females are more likely to be fertile than males. Assume that children are
born to a female with Down syndrome and a normal 46-chromosome male. What
proportion of the offspring would be expected to have Down syndrome?
1. One half of the offspring will be expected to have Down syndrome.
2. None of the offspring would be expected to have Down syndrome.
3. All the children would be expected to have Down syndrome.
4. Two-thirds of the offspring would be expected to have Down syndrome.
15.
While the most frequent forms of Down syndrome are caused by a random
nondisjunction of chromosome #21, Down syndrome occasionally runs in families with
apparently normal individuals having children with Down syndrome. The cause of this
form of familial Down syndrome is
1.
2.
3.
4.
16.
an inversion involving chromosome #21
too many X chromosomes
a translocation between chromosome #21 and another chromosome
a maternal age effect
Nondisjunction
1. can lead to aneuploidy.
2. does not alter a karyotype.
3. turns a diploid genome into a haploid genome.
4. is responsible for familial Down syndrome.
17.
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
1. chaos.
2. inversions.
3. incomplete penetrance.
4. position effect.
18.
The genetic material of most living things is
1. deoxyribonucleic acid.
2. ribonucleic acid.
3. RNA.
4. polysaccharide.
19.
In the classic experiment conducted by Hershey and Chase, why were the bacteria
radioactive after they had been infected with viruses which had been grown on medium
containing 32P?
1. The bacteria had incorporated radioactive proteins into their
cell membranes.
2. The radioactive viruses (coats plus DNA) were injected into the bacteria.
3. The bacteria contained radioactive viral DNA
4. None of the above.
20.
The basic structure of a nucleotide includes the following components:
1. amino acids.
2. base, sugar, phosphate.
3. phosphorous and sulfate.
4. all of the above.
21.
If 15% of the nitrogenous bases in a sample of DNA from a particular organism is
thymine, what percentage should be cytosine?
1. 15%
2. 30%
3. 70%
4. 35%
22.
Is the molecule to the right DNA or RNA, and is the arrow closest to
the 5' or 3' end?
1. DNA, 3’ end
2. RNA, 3’ end
3. DNA, 5’ end
4. RNA, 5’ end
A
T
Refer to the following figures for question 23
a
d.
23.
b.
c.
e.
Which diagram shows a nucleotide with a purine base?
1. Figure a
2. Figure b
3. Figure c
4. Figure d
24.
In E. coli bacteria, which terms accurately reflect the nature of replication of the
genome?
1. bidirectional and fixed point of initiation.
2. unidirectional and conservative.
3. unidirectional and fixed point of initiation.
4. multirepliconic.
25.
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?
1. 100% of the molecules were of heavy density.
2. 50% were of heavy density, 50% were intermediate density.
3. 100% were of intermediate density.
4. 50% were of light density, 50% were intermediate density.
For questions 26-27 refer to the replication fork depicted below:
5’
3’
B
A
3’
5’
B
B
C
5’
3’
26.
Item A is called
1.
2.
3.
4.
27.
the leading strand.
primase.
single strand binding protein.
an Okazaki fragment.
B is most likely:
1. an RNA primer.
2. a DNA primer.
3. DNA polymerase III.
4. an exonuclease.
28.
The enzyme called “telomerase”
1. contains a reverse transcriptase.
2. contains a molecule of RNA.
3. helps regulate the cell cycle.
4. all of the above.
29.
The enzyme that appears to be inappropriately expressed in cancer cells and
helps to maintain the very ends of eukaryotic chromosomes is called
1. topoisomerase.
2. ligase.
3. telomerase.
4. terminase.
30.
In addition to highly repetitive and unique DNA sequences, a third category of
DNA sequences exists. What is it called and what types of elements are involved?
1.
2.
3.
4.
composite DNA, telomeres and heterochromatin
dominant DNA, euchromatin and heterochromatin
multiple gene family DNA, hemoglobin and 5.0S RNA
moderately repetitive DNA, SINEs, LINEs, and VNTRs
31.
Chromatin of eukaryotes is organized into repeating interactions with protein
octomers called nucleosomes. Nucleosomes are composed of which class of
molecules?
1.
2.
3.
4.
32.
histones
glycoproteins
lipids
nonhistone chromosomal proteins
In eukaryotic chromatin, the next highest level of organization above "beads-ona-string" is
1. the 30 nm fiber.
2. chromatin loops.
3. the nuclear matrix.
4. the metaphase chromosome.
33.
Most DNA occurring in nature
1.
2.
3.
4.
is relaxed.
is supercoiled.
is “naked” and is not complexed with proteins
has ends that are free to rotate.
34.
Mutations which arise in nature from no particular artificial agent are called
1. induced mutations.
2. spontaneous mutations.
3. chromosomal aberrations.
4. cosmic mutations.
35.
Two terms used to describe categories of mutational nucleotide substitutions in
DNA are called
1. base analogues and frameshift.
2. error prone and spontaneous.
3. transversions and transitions.
4. euchromatic and heterochromatic.
36.
A class of mutations that results in multiple contiguous amino acid changes in
proteins is likely to be which of the following?
1. base analogue.
2. transversion.
3. transition.
4. frameshift.
37. Ultraviolet light causes pyrimidine dimers to form in DNA. Some individuals are
genetically incapable of repairing some dimers at "normal" rates. Such individuals are
likely to suffer from
1. xeroderma pigmentosum.
2. phenylketonuria.
3. muscular dystrophy.
4. Huntington disease.
38.
In mammals, DNA double-strand breaks may be repaired by
1. mismatch repair.
2. base excision repair.
3. nucleotide excision repair.
4. nonhomologous end-joining
39.
In the first step of base excision repair
1.
2.
3.
4.
40.
a phosphodiester bond is broken
two nicks are made surrounding the DNA damage
a break in DNA is sealed by DNA ligase
a base is removed by the action of a DNA glycosylase
The simplest type of transposable elements in bacteria are known as
1. transposons.
2. insertion sequences.
3. Ds elements.
4. Ty elements.
That’s all!
Answer key
ITEM NUMBER
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
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40
CORRECT ANSWER
3
4
3
2
2
1
1
4
3
2
2
2
2
1
3
1
4
1
3
2
4
1
4
1
4
4
1
4
3
4
1
1
2
2
3
4
1
4
4
2