Sample Examination Questions for Exam 2 Material Warning!

Sample Examination Questions for Exam 2 Material
Biology 3300 / Dr. Jerald Hendrix
Warning! These questions are posted solely to provide examples of past test questions.
There is no guarantee that any of these questions will be on any examination in the future.
Students are responsible for all of the material covered in lectures, assigned readings,
textbook problems, laboratories, and any other assigned work. Since these samples have
been taken from several past exams, some questions may be very similar or identical. On
short answer, essay questions, and genetics problems, the point values from previous
exams have been included to give an indication of approximately how much “weight”
was given to a question in the past; however, there is no guarantee that any particular
question, format, or point distribution will be used on any examination.
The following information pertains to questions 3 and 4.
The two strands of a double-stranded DNA molecule can be separated by heating the
strands. The temperature at which the strands separate is called the melting point of the
DNA. A DNA molecule with a larger number of hydrogen bonds between the strands will
have a higher melting point than a DNA molecule with fewer hydrogen bonds. In
questions 3 and 4:
A = the number of deoxyadenosine nucleotides in a DNA molecule
C = the number of deoxycytidine nucleotides in a DNA molecule
G = the number of deoxyguanosine nucleotides in a DNA molecule
T = the number of deoxythymidine nucleotides in a DNA molecule
3.
The value of (G + C)/(A + T)
(a)
(b)
(c)
(d)
(e)
is equal to one if the DNA is single-stranded.
is equal to one is the DNA is double-stranded.
is larger in DNA samples having a higher melting point.
is lower in DNA samples having a higher melting point.
none of the above.
GC base pairs have 3 hydrogen bonds; AT base pairs have only 2. Therefore, if a
sample of DNA has more GC, then we expect it to have a higher melting point.
4.
The value of (A + C)/(G + T)
(a)
(b)
(c)
(d)
(e)
is equal to one if the DNA is single-stranded.
is equal to one is the DNA is double-stranded.
is larger in DNA samples having a higher melting point.
is lower in DNA samples having a higher melting point.
none of the above.
In double-stranded DNA, the amount of A equals T, and C equals G, so (A+C)/(G+T)
equals one.
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6.
Which of the following enzymes is used to cleave vector and chromosomal DNA
in the formation of recombinant DNA molecules?
(a)
(b)
(c)
(d)
(e)
7.
In a recombinant DNA expression library
(a)
(b)
(c)
(d)
10.
trisomy
monosomy
tetrasomy
chromosomal deletion
chromosomal translocation
A bacterial enzyme that recognizes a specific nucleotide sequence on doublestranded DNA, then cleaves the DNA at the recognition site, is called a
(a)
(b)
(c)
(d)
(e)
16.
the cloned DNA is inserted into an RNA vector to facilitate expression.
the cloned DNA is inserted into a plasmid vector at a position adjacent to a
bacterial promoter region to allow for transcription and subsequent gene
expression.
the cloned DNA is bound directly to the small ribosome subunit which
results in expression in about 85% of the colonies isolated.
the cloned DNA is not expressed inside the bacterial cell.
A human being with a chromosome number of 47 in each somatic cell has which
of the following conditions?
(a)
(b)
(c)
(d)
(e)
12.
RNA polymerase
aminoacyl tRNA synthetase
Type I restriction endonuclease
Type II restriction endonuclease
reverse transcriptase
Type I restriction endonuclease.
Type II restriction endonuclease.
modification methylase.
DNA polymerase.
5' exonuclease.
The map distance calculated in a two-factor test cross is most likely to be an
underestimate of the actual distance due to
(a)
(b)
(c)
(d)
(e)
epistasis.
nondisjunction.
interference.
codominance.
double crossovers.
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17.
Which of the following enzymes can be used to synthesize cDNA from an mRNA
template?
(a)
(b)
(c)
(d)
(e)
23.
DNA-directed RNA polymerase
DNA-directed DNA polymerase
restriction endonuclease
DNA ligase
reverse transcriptase
The oocytes of the frog Xenopus contain lampbrush chromosomes. These are
large, visible chromosomes with both condensed regions and with greatly
extended regions that resemble the bristles on a lampbrush. If Xenopus oocytes are
incubated with radiolabeled uridine for a brief period of time and
autoradiographed, only the extended regions are radiolabeled. What does this
indicate?
(a)
(b)
(c)
(d)
(e)
DNA is replicated repeatedly without mitosis in the extended regions.
Protein synthesis is occurring in the extended regions.
Transcription is occurring in the extended regions.
Crossing-over is occurring in the extended regions.
Assortment is occurring in the extended regions.
Uridine is found in RNA, so radiolabeled uridine would serve as a label for
transcripition, or RNA synthesis.
25.
Which of the following terms best describes a ribose molecule with a cytosine
molecule attached to its 1' carbon and an -OH group attached to its 5' carbon?
(a)
(b)
(c)
(d)
(e)
31.
a nitrogenous base
a nucleoside
a nucleotide
a base pair
a promoter
In a karyotype of a man whose wife has had several miscarriages, it was
discovered that he had only 45 chromosomes. However, he displayed no obvious
phenotypic defects. Furthermore, one of the chromosomes of pair 15 was
abnormally long. Which of the following terms best describes the condition of this
man?
(a)
(b)
(c)
(d)
(e)
monosomy
trisomy
tetrasomy
translocation carrier
cri-du-chat
Page 3
32.
Which of the following terms best describes the procedure in which DNA
fragments are separated on an agarose gel, transferred onto a nylon membrane,
and screened with a labeled hybridization probe?
(a)
(b)
(c)
(d)
(e)
1.
dot blotting
dash blotting
Southern blotting
Northern blotting
Western blotting
In the tomato plant, round fruit shape (R) is dominant over long shape (r). Smooth
fruit skin (P) is dominant over peachy skin (p). The genes for fruit shape and fruit
texture are linked on the same chromosome.
A heterozygous round, heterozygous smooth plant was crossed with a long,
peachy plant. The results are given in the table below.
Smooth round
Smooth long
Peachy round
Peachy long
39
463
451
47
(a)
In the heterozygous parent plant, was the linkage between the genes cis or trans?
(2 pt)
trans
(b)
Calculate the map distance between the genes. (2 pt) 8.6 map units
(c)
Is the distance calculated in (b) more likely to be an overestimate or an
underestimate of the true distance between the genes? Briefly explain your
answer. (4 pt) An underestimate, due to the possibility of double crossovers
2.
If the value of interference in a three factor testcross is negative, what does this
signify with respect to the process of recombination between the chromosomes in
the homologous pair? (4 pt)
A negative interference value indicates that the first crossover has somehow
increased the probability of a second crossover occurring.
5.
How can chromosomal nondisjunction during meiosis result in an offspring with
trisomy or monosomy? (4 pt)
Nondisjunction is the failure of homologous chromosomes to separate during
meiosis I, or sister chromatids to separate during meiosis II. The result is an
abnormal gamete that either contains two copies of a chromosome instead of
one, or doesn’t contain a copy at all. If an abnormal gamete containing two
copies of a chromosome is fertilized with a normal gamete containing one copy,
the result is a zygote with a trisomy for that chromsome. If an abnormal gamete
missing a chromosome is fertilized with a normal gamete, the result is a
monosomy.
Page 4
6.
Briefly distinguish between trisomy Down’s syndrome and translocation Down’s
syndrome. (4 pt)
In trisomy Downs syndrome, the individual has three copies of chromosome 21,
for a total of 47 chromosomes. In translocation Downs syndrome, the
individual has a total of 46 chromosomes, with two normal 21 chromosomes,
one normal 14, and an abnormal 14/21. These individuals would be both show
the phenotypes for Downs syndrome: retardation, heart problems, facial effects,
etc. A translocation Downs carrier shows no symptoms of Downs syndrome, but
often will have children with the syndrome. The translocation carrier has 45
chromosomes, with one normal 21, one normal 14, and one abnormal 14/21.
7.
What are polytene chromosomes? How are they formed in the cells of
Drosophila? What feature of polytene chromosomes suggests that the chromatin
of the interphase nucleus is folded in an ordered fashion? (6 pt)
Not covered.
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3.
In maize, the gene for colored kernels (C) is dominant over its recessive allele for
colorless kernels (c). The gene for full kernels (Sh) is dominant over its allele for
shrunken kernels (sh). The gene for nonwaxy endosperm (Wx) is dominant over
its allele for waxy endosperm (wx). These three genes are located on the same
chromosome.
A plant heterozygous for all three genes was crossed with a plant that was
homozygous recessive for all three genes. The progeny are given in the table
below.
colored
colored
colored
colored
colorless
colorless
colorless
colorless
shrunken
shrunken
full
full
shrunken
shrunken
full
full
waxy
nonwaxy
waxy
nonwaxy
waxy
nonwaxy
waxy
nonwaxy
305
112
74
22
18
66
128
275
(a)
Determine the sequence of these three genes. (2 pt) Sh – C – Wx
(b)
State the type of linkage (cis or trans) that existed between each pair of genes in
the heterozygous parent. (3 pt)
Sh – C: trans; C – Wx: trans; Sh – Wx: cis
(c)
Calculate the map distance between each pair of genes. (4 pt)
Wx – C: 28 map units
Sh – C: 18 map units
(d)
Calculate the coefficient of coincidence and the interference of the cross. (4 pt)
Coefficient of coincidence = 0.794
Interference = 0.206
10.
A geneticist is studying the structure of two different viruses, A and B, each of
which contains DNA. The nucleotide base composition of each virus is given in
the table below.
Base
Adenine
Guanine
Cytosine
Thymine
Amount of the base in
Virus A
Virus B
10000
7000
15000
18000
15000
10000
10000
15000
Based on this data, what differences do you predict will exist between the
structures of these two viruses? (4 pt) Virus A most likely has double-stranded
DNA, but virus B has single-stranded DNA.
Page 6
11.
Give two reasons why type II restriction endonucleases are useful in preparing
recombinant DNA molecules. In your answer, you should contrast type II
restriction endonucleases with type I restriction endonucleases. (4 pt)
1. Type II restriction endonucleases recognize specific nucleotide sequences
and cut the DNA at the recognition site, making reproducible fragments. This is
different from type I enzymes, because type I enzymes cut DNA at a random
distance from teh recognition site and make random fragments, so they are not
useful for recombinant DNA technology.
2. Many Type II restriction endonucleases have palindromic recognition
sequences and make staggered cuts, so they make short single-stranded
complementary ends (“sticky ends”). This means that if DNA from two
different sources is cut with the same Type II enzyme, the ends can be spliced
together (the DNA is sealed with another enzyme called DNA ligase), making
recombinant DNA.
1.
Give the correct name for each of the following structures. (4 pt each)
O
H
HO
NH 2
H
H3C
O
N
C H
N
O
HO
O
H
N
O C H
P
O
N
OH
(a)
OH H
deoxythymidine
(b)
O
N
(c)
N
N
H
guanine
OH H
deoxycytidine monophosphate
H
N
H
HO C H
O
NH 2
OH
(d)
ribose
Page 7
OH
OH
O
O
NH2
O
N
H
HO C H
O
N
HO
O
P
H
N
H
O C H
O
N
O
OH
(e)
OH H
deoxycytidine
2.
Briefly describe the B-helix structure of double-stranded DNA. Include its length
in Angstroms (Å) per turn, the number of bases per turn, its grooves, its basepairing, and the orientation of the strands with respect to each other. (10 pt)
(f)
OH OH
uridine monophosphate
The sugar-phosphate backbones of the strands are wound around the axis of
the helix, forming two grooves that spiral around the axis: a large groove, or
major groove, and a smaller groove, or minor groove. The bases are stacked in
the center of the helix, where they pair according to the base pairing rules: A to
T, with two hydrogen bonds, and C to G with three hydrogen bonds. The helix
is 20 Å in diameter, 10 bases per turn, 34 Å per turn, 3.4 Å per base. The
strands are antiparallel, meaning that one strand is in the 5’ → 3’ orientation,
and the other strand is 3’ → 5’.
3.
Briefly explain the two major lines of evidence used by Watson and Crick to
deduce the structure of the B-DNA helix. (6 pt)
X-ray diffraction crystallography of DNA fibers showed that DNA was in the
form of a helix, and from the diffraction pattern Watson and Crick were able to
calclulate the dimensions of the helix. Chargaff’s data on base composition of
DNA showed that the A:T and C:G molar ratios were 1:1, suggesting that the
DNA was a double helix with A-T and C-G base pairs.
5.
What is bacterial competence? (2 pt)
Competence is the ability of bacterial cells to take up isolated DNA molecules
by transformation. Some bacteria are naturally competent for transformation,
but other bacteria, such as E. coli, must have competence induced by treatment
with calcium chloride (CaCl2)
Page 8
12.
You wish to study the gene in humans for the protein actin. You know that this
protein is made in cultured human fibroblast cells, but it is not the only protein
made in the cells. You also know the amino acid sequence of human actin. At the
time you are doing your experiment, no actin genes from other species have been
isolated. You have available to you a human recombinant DNA library, cultured
fibroblasts, purified antibodies specific for the protein actin, and all the supplies
and equipment needed to perform the techniques that we discussed in class. The
recombinant library is not an expression library. Outline two different
experimental approaches to isolate the actin gene from the recombinant library.
(14 pt)
Since the library is not an expression library, you’ll need a probe to be able to
screen the colonies in your library for the actin gene. Where can you obtain
your probe? There are two ways we could do this:
1. Isolate the mRNA for actin from cultured human fibroblast cells, and use the
mRNA to make a cDNA probe using reverse transcriptase.
2. Create a synthetic DNA probe by deducing possible sequences of the actin
gene using the known amino acid sequence of the protein actin and the genetic
code table.
You can’t use a gene from an existing species since, according to the problem,
non exist at the time you are doing your work. Also, you cannot screen the
colonies in the library for expression since it is not an expression library (i.e.,
the fragments are not cloned in an expression vector).
Page 9
13.
Sickle-cell anemia is caused by a mutation in the human β-globin gene. The three
possible genotypes are homozygous for normal
β -globin, heterozygous carrier (having both the normal and sickle-cell genes),
and homozygous for sickle cell anemia.
Recombinant DNA technology has been used as the basis for prenatal diagnosis of
sickle cell anemia. In a very high percentage of the cases observed, the normal
human β-globin gene is carried on a 7600 bp human DNA fragment from a HpaI
digest, while the sickle-cell gene is carried on a 13000 bp HpaI fragment. HpaI is
a type II restriction endonuclease.
You have available to you:
• A radiolabeled sample of recombinant DNA consisting of a bacterial plasmid
vector (4000 bp) carrying the 7600 bp HpaI fragment from the normal human
genome.
• A sample of chromosomal DNA from each member of a couple thought to be
carriers of the sickle-cell trait and expecting their first child.
• A sample of the chromosomal DNA obtained from the fetal cells in the
amniotic fluid from the uterus of the pregnant woman.
• All of the supplies and equipment necessary to perform the techniques that we
discussed in class.
How can you tell if the parents are heterozygous carriers of the sickle-cell trait,
and if the fetus is normal, a heterozygous carrier, or homozygous for sickle-cell
anemia? (12 pt)
A helpful hint: the genes for normal and sickle-cell β-globin share sufficient
sequence homology to hybridize with each other.
For each parent’s DNA and the fetal DNA, cut the DNA with HpaI and
separate the fragments by agarose gel electrophoresis. Using the Southern
blotting technique, transfer the fragments to a nylon (or nitrocellulose)
membrane, and probe the membrane with the radiolabeled 7600 HpaI
fragment.
Expected results:
Homozygous normal β-globin:
Homozygous sickle cell anemia:
Heterozygous sickle cell carrier:
11.
A single band at 7600 bp
A single band at 13000 bp
Two bands, one at 7600 and one at 13000
Distinguish between trisomy Downs syndrome and translocation Downs
syndrome. (5 pt)
Duplicate question
Page 10
9.
In the pea plant, the gene for purple flower color is dominant to its allele for red
flower color. The gene for long pollen grain shape is dominant to its allele for
round pollen grain shape. These genes are located on the same chromosome. In a
two factor testcross experiment, a plant that was heterozygous for both genes was
crossed with a plant having red flowers and round pollen grains. The following
results were obtained.
445 Purple flower, long pollen
48 Purple flower, round pollen
52 Red flower, long pollen
455 Red flower, round pollen
(a)
In the heterozygous parent used to make the cross, was the linkage between the
genes in the cis or the trans configuration? (2 pt) cis
(b)
What is the map distance between the genes for flower color and pollen shape?
(4 pt) 10 map units
(c)
Is the distance that you calculated in part (a) most likely an overestimate or an
underestimate of the actual map distance? Why? (4 pt) Underestimate, due to
double crossovers
10.
In corn, the gene for colored kernels (C) is dominant over its allele for colorless
kernels (c). The gene for full kernels (Sh) is dominant over shrunken (sh). The
gene for nonwaxy endosperm (Wx) is dominant over waxy (wx). These genes are
located on the same chromosome. A plant heterozygous for all three traits was
crossed with a triply homozygous recessive plant, and the following progeny were
obtained.
colored
colorless
colorless
colored
colorless
colored
colored
colorless
shrunken
full
shrunken
full
shrunken
full
shrunken
full
waxy
waxy
waxy
waxy
nonwaxy
nonwaxy
nonwaxy
nonwaxy
305
128
18
74
66
22
112
275
(a)
What is the correct order of these genes on the chromosome? (3 pt)
Duplicate question
(b)
What are the linkages (cis or trans) between each pair of genes in the
heterozygous parent? (3 pt)
(c)
What are the distances in map units between each pair of genes? Show your
work. (4 pt)
(d)
What is the coefficient of coincidence and the interference? (4 pt)
Page 11