Genetics Essentials: final exam review previous exam Flashcards

scalloped (sd) is an X-linked recessive and ebony (e) is an autosomal
recessive mutation. What proportion of scalloped, ebonyfemales
(relative to whole population) is expected in the F2 starting with a
true breeding scalloped female which is wild type forebony
mating with a true breeding male mutant only for ebony.A.
0B. 1/32C. 1/16 D. 1/8E. 3/16

C

A husband and wife have normal vision, although both of their fathers
are colorblind. What is the probability that their first child will be
a daughter who is colorblind?A. 0 B. 1/8C.
1/4D. 1/2E. 1

A

The following outcomes are noted for matings of Martian snapdragons.
Green x Green = Green; Orange X Orange = Orange;Orange X Green =
Purple; Purple X Purple = 1/4 Green, 1/2 purple, 1/4 orange. This is
most likely due toA. X-linked genesB. epistasisC.
multiple alllelesD. CodominanceE. Incomplete Dominance

E

Which of the following is the result of dosage compensation in human
femalesA. MLE bodyB. Turner bodyC. Barr Body
D. Lyon bodyE. Chromatin body

C

If a rare genetic disease is inherited on the basis of an X-linked
gene, one would expect to find which of the following:A.
Affected fathers have 100% affected sonsB. Affected fathers have
100% affected daughtersC. Carrier mothers have 50% affected sons
D. Carrier mothers have 50% affected daughtersE. Carrier
mothers have 100% affected sons

C

A drosophila with 4 X chromosomes, a Y chromosome and 3 sets of
autosomes will be aA. normal maleB. normal femaleC.
metafemale D. intersexE. metamale

C

What part of the Y chromosome allows it to pair with the X chromosome
in meiosis?A. PARB. SRYC. TDFD. PTAE. MSY

A

B is a dominant giving black color. b is the recessive allele causing
blue color. In a population of 100 B/b individuals, four
blueanimals are seen. This is an example ofA. incomplete
dominanceB. partial dominanceC. reduced penetranceD.
reduced expressivityE. poor record keeping

C

Shell orientation in snails is due to a maternal effect gene. A true
breeding sinistral (recessive) is crossed to a true
breedingDextral (dominant). The offspring from that cross are
self-crossed. What will be the expected ratio of shell types?A.
All sinistralB. All Dextral C. Half sinistral, half
DextralD. � Dextral, � sinistralE. � sinistral, � Dextral

B

A couple has five children. What is the probability that they would
have five boys?A) 1/2B) 1/4C) 1/8D)
1/16E) 1/32

E

In peas, axial (A) flower position is dominant to terminal (a), tall
(L) is dominant to short (l), and yellow (Y) is dominant togreen
(y). If a plant that is heterozygous for all three traits is allowed
to self-fertilize, how many of the offspring would be
dominantfor all three traits?A) 3/64B) 9/64C)
27/64 D) 32/64E) 64/64

C

Contact points between nonsister chromatids that mark the locations
of DNA-strand exchange are calledA) synaptonemal
complex.B) metaphase plate.C) chiasmata. D)
kinetochore.E) centrosome.

C

A chromosome with a centromere at the end is calledA.
metacentricB. submetacentricC. acrocentricD.
telocentric E. centrocentric

D

Morgan was the first scientist to analyze the white-eyed male
Drosophila mutation which helped explainA) autosomal
dominance.B) random mutation.C) crossing over.D)
independent assortment.E) X-linked inheritance.

E

The product H substance is needed to express the blood antigen. A
mating of IA IB Hh X IA IB hh should produce what ratio ofblood
types?a) 1/4 AB, 1/8 A, 1/8 B, 1/2 O b) 3/8 AB, 3/16 A,
3/16 B, 1/4 Oc) 1/4 AB, 3/16 A, 3/16 B, 3/8 Od) 3/16 AB,
3/8 A, 3/16 B, 1/4 Oe) 1/8 AB, 1/4 A, 1/8 B, 1/2 O

A

If a common slipper limpet Crepidula fornicata lands on top of a
female limpet, it willA. become a male and cannot changeB.
become a female and cannot changeC. become a male and change to
female if another limpet lands on top of it D. become a female
and change to a male if there are too many femalesE. beg the
limpets pardon and move to a different part of the ocean

C

True breeding white flowering plants are crossed to true breeding red
flowering plants. The offspring are red flowered. Whenthese
flowers are self crossed, the resultant offspring are 87 red, 32 pink
and 39 white flowered plants. Which of the followingpathways is
most likely the cause of thisA. white
---A--->pink---B--->redB. pink
---A--->pink---B--->redC. red<---A---white
---B--->redD. white ---A--->pink white---B--->pink both
genes producing pink together give red

A

If a trait is X-linked recessive, who would express the
trait?A) homozygous dominant females and hemizygous recessive
malesB) heterozygous recessive females and hemizygous dominant
malesC) homozygous recessive females and hemizygous recessive
males D) heterozygous dominant females and hemizygous dominant
malesE) the same proportions of females and males

C

If purple (P) is dominant to white (p) and axial (A) is dominant to
terminal (a), and in a cross of white, axial to purple, axial
theratio of offspring is6/16 white, axial;2/16
white, terminal;6/16 purple, axial;2/16 purple,
terminal,then what is the genotype of the parents?A. PPAA
X ppaaB. PpAa x PpAaC. PpAa X PpaaD. PpAa X
ppAaE. Ppaa X Ppaa

D

Color blindness is X-linked recessive and blood type is autosomal. If
two parents who are both Type A and have normal visionproduce a
son who is color blind and type O, what is the probability that their
next child will be a female who has normal visionand is type
O?A. 0B. 1/8C. 1/4D. 1/2E. 1

B

26) Which of the following occurs only in Meiosis II:A.
Separation of homologuesB. Synaptonemal complex
formationC. Formation of spindle polesD. Separation of
centromeres of sister chromatidsE. None of the above

D

A non-color blind father is married to a woman who is also not color
blind. They have four sons, two of whom are color blind.They
have a child with Klinefelter syndrome. This child suffers from color
blindness. This occurred because of a normal gametefusing with a
gamete which is the result of nondisjunction in:A. Either
Meiosis I or II in maleB. Meiosis II in maleC. Either
Meiosis I or II in femaleD. Meiosis I in femaleE. Meiosis
II in female
What is the sex of the color blind child in the
previous questionA. male B. femaleC. intersex

E, A

In a cross of AaBbCcDdEeFf X AaBbccDdEeFf, what proportion will have
the ABCDeF phenotype?A. 27/64B. 27/128C.
27/512D. 81/512E. 81/2048

E

Mendel crossed purebred wrinkled, green-seeded plants with purebred
round, yellow-seeded plants. The F1 progeny were
selfcrossed.Which of the following numbers are most likely to
result from this cross.A. 22 round, yellow; 26 round, green; 28
wrinkled, yellow, 24 wrinkled greenB. 91 round, yellow; 32
round, green; 29 wrinkled, yellow, 8 wrinkled green C. 64 round,
yellow; 26 round, green; 27 wrinkled, yellow, 27 wrinkled
greenD. 11 round, yellow; 26 round, green; 27 wrinkled, yellow,
87 wrinkled green
What is the chi-square value of the product of the
correct cross from problem 30 precedingA. 2/5B.
4/5C. 7/25D. 49/90E. 26/45
32) What is the probability of this event occurring,
chi-square table followsA. >0.9 B. 0.9>0.5C.
0.5>0.1D. 0.1>0.05Degrees of Freedom 0.9 0.5 0.1
0.05===========================================================1
0.02 0.46 2.71 3.842 0.21 1.39 4.60 5.993 0.58 2.37 6.25
7.824 1.06 3.86 7.78 9.49

B, E, A

If a male bird that expresses a recessive Z-linked mutation is
crossed to a wild type female, what proportion of the
totalprogeny will be mutant females?A. 0%B.
25%C. 50% D. 75%E. 100%

C

Pudgy Birds come in two forms, listless and active. Independent of
who they mate with, listless mothers always give birth
tolistless offspring and their daughters always give birth to
more listless offspring of both sexes. This is most likely due
to:A. Sex-limited traitB. Recessive lethalC.
Mitochondrial genesD. Maternal effect genes

C

True breeding white flowering plants are crossed to true breeding red
flowering plants. The offspring are red flowered. Whenthese
flowers are self crossed, the resultant offspring are 92 red, 57 pink
and 12 white flowered plants. Which of the followingpathways is
most likely the cause of thisA. white
---A--->pink---B--->redB. pink
---A--->pink---B--->redC. red<---A---white
---B--->redD. white ---A--->pink white---B--->pink both
genes producing pink together give red

D

If a rare genetic disease is inherited on the basis of an Y-linked
gene, one would expect to find which of the following:A.
Affected fathers have 100% affected sonsB. Affected fathers have
100% affected daughtersC. Carrier mothers have 50% affected
sonsD. Carrier mothers have 50% affected daughtersE.
Carrier mothers have 100% affected sons

A

The correct order in mitosis is:A. Methaphase, Interphase,
Telophase, Anaphase, ProphaseB. Interphase, Telophase,
Metaphase, Prophase, AnaphaseC. Interphase, Prophase, Metaphase,
Anaphase, TelophaseD. Telophase, Metaphase, Interphase,
Prophase, Anaphase

C

A cross of two heterozygotes yields a 2:1 ratio of phenotypes. This
is indicative of:A. Incomplete dominanceB.
codominanceC. Recessive lethalD. dominant lethalE.
multiple alleles

C

A carrier mother with normal vision and a colorblind father have a
child who is XXXXYY and colorblind. This is the result of:A.
Nondisjunction in Meiosis I and II in the father and Meiosis I in the
motherB. Nondisjunction in Meiosis I in the father and
motherC. Nondisjunction in Meiosis I in the father and Meiosis I
and II in the motherD. Nondisjunction in Meiosis II in the
father and Meiosis I and II in the mother.E. Nondisjunction in
Meiosis I and II in the father and Meiosis II in the mother.

E

During meiosis I, when does homologous chromosome pairing and
recombination occur?A) prophase IB) pro-metaphase
IC) metaphase ID) anaphase IE) telophase I

A

1) If you are given a recombination frequency of 34% between genes X
and Y and 27% between X and Z, can you predict the order ofthe
three genes?A) Yes; the order is X-Z-Y.B) Yes; the order
is X-Y-Z.C) Yes; the order is Z-X-Y.D) No; based on this
data alone, the order could be Z-Y-X or X-Y-Z.E) No; based on
this data alone, the order could be X-Z-Y or Z-X-Y.

E

Which of the following has the largest overall affect on the allele
frequency of a population?A. mutationB. migrationC.
genetic driftD. inbreeding

C

In a population of 20,000 individuals, 200 men are afflicted with a
recessive, X-linked disease. How many woman would beexpected to
be afflicted in this population?A. 1B. 2C. 4D.
8E. 24

C

The clearing made by bacteriophages in a "lawn" of bacteria
on an agar plate is called a ________.A) clear zoneB)
lysogenic zoneC) prophageD) plaqueE) host range

D

Heritability of corn height is 0.5. Farmer Bill's crop averages 8 ft
tall. He takes all of his 10 foot tall plants and crosses them.
Theexpected average height of the resultant offspring
is:A. 8 ftB. 8.5 ftC. 9 ftD. 9.5 ftE. 10 ft

C

Assume that a cross is made between AaBb and aabb plants and that the
offspring occur in the following numbers: 106 AaBb, 48Aabb, 52
aaBb, 94 aabb. These results are consistent with the following
circumstance:A) sex-linked inheritance with 30% crossing
over.B) linkage with 50% crossing over.C) linkage with
approximately 33 map units between the two gene loci. D)
independent assortment.E) 100% recombination.

C

Outbreeding in a population tends to:A. select for dominant
allelesB. decrease heterozygotesC. eliminate recessive
allelesD. decrease homozygotes

D

If 4% of individuals in a population express a recessive gene, what
percentage of the total population are heterozygous for
thegene?A. 4%B. 10%C. 16%D. 32%E. 64%

D

14) The following progeny phenotypes result from a cross of an A/a
B/b individual with an a/a b/b individualA B 300a b
300A b 200a B 200What is the recombination rate
between A and B?A. 0.1B. 0.2C. 0.3D. 0.4
E. 0.5

D

An allele affecting antler size has a frequency of .25 in one group
of caribou, but is not found in another group living in a
nearbymountain valley. This is an example of:A.
inbreedingB. mutationC. genetic drift D. gene
flowE. selection

C

Three years later, the same caribou groups are surveyed and both now
have a frequency of .20 for the antler size allele. This is
anexample of:A. inbreedingB. mutationC.
genetic driftD. migration E. selection

D

In a population of people, 8% of the males are red/green color blind
(X-LINKED). Assuming the population is in equilibrium,what
proportion of the females would be expected to be afflicted?A.
0.64% B. 4%C. 64%D. 6.4%E. 0.32%

A

In which form of bacterial recombination does the bacteria find and
take up the DNA instead of having it injected into the cell?A.
ConjugationB. InterferenceC. Transformation D.
TransductionE. Liposuction

C

Cheetahs are an extreme example ofA. mutationB.
migrationC. genetic drift D. inbreeding

C

Villagers in isolated villages in the mountains tend to have higher
rates of recessive genetic diseases because of:A. lower ozone
levels at high altitudesB. less diverse dietsC. increased
inbreeding D. no cable TV

C

Natural selection:A) eliminates alleles from a
populationB) has no affect on allele frequency C)
increases heterozygosityD) alters allele frequencies, but
usually does not remove alleles from a populationE) increases homozygosity

B

In a population of 10,000 individuals, 100 express a recessive trait.
How many heterozygotes are there expected to be?A. 600B.
1,000C. 1,800 D. 2,400E. 3,000

C

The power of bacterial genetics isA. Large number of genes so
mapping can be done more completelyB. Many varied phenotypes so
markers are easy to determineC. Haploid genome makes genetics
easier to carry outD. Very small organisms are easier to
manipulateE. Very fast generation time and large numbers of offspring

E

Why is mutation critical for change in the allele frequencies of a
populationA. It creates the new mutations that are needed for
variationB. It drives the change of frequency of allelesC.
It can create the multiple alleles that are needed for true
variationD. It stops alleles from going into fixation by
recreating lost alleles

A

Which of the following is not true of translocations?A. They
can cause speciation to occurB. They usually alter expression of
multiple genes C. They can bring the control region from one
gene next to a different geneD. They reduce the number of viable
gametes produced

B

Among the human autosomes, there are 3 viable trisomy conditions, all
with developmental problems, but no viable monosomyconditions.
This is because:A. Trisomies alter the balance of developmental
signals and monosomies uncover recessive lethalsB. Both
trisomies and monosomies uncover recessive lethalsC. Monosomies
alter the balance of developmental signals and trisomies uncover
recessive lethalsD. Both trisomies and monosomies alter the
balance of developmental signals

A

In generalized transduction, a phage introduces a segment of donor
DNA into the recipient cell. This is followed byrecombination of
the donor fragment with the recipient chromosome. Which of the
following must occur?A) circularization of the donor fragment
before recombinationB) a pair of crossovers between the donor
segment and the recipient C) degradation of one of the two
strands of phage genomeD) replication of the donor segment
before recombinationE) a single crossover between the donor
segment and the recipient chromosome

B

If you wanted to release negative supercoiling in a DNA chromosome,
which of the following would you mix with the DNA?A.
ShelterinB. HistonesC. complementary DNA strandsD. topoisomerase

D

If a DNA molecule of 50 base pairs contains 15 cytosine bases (C),
how many thymine bases will it have?A. 10B. 15C.
30D. 35 E. 50

D

Which of these sequences could form a hairpin??. 5?
GGGGTTTTCCCC 3? ?. 5? AAAAAAAAAAAA 3??. 5? ACACACACACAC
3?D. 5? TTTTTTCCCCCC 3

A

Which activity is not associated with DNA polymerases?A.
Ability to �read� a template and incorporate appropriate nucleotides
in the growing strandB. 5� to 3� synthesis of the new
strandsC. Ability to initiate DNA synthesis on a completely
single stranded DNA molecule D. 3� to 5� exonucleaseE. Proofreading

C

After sigma factor guides the RNA polymerase to the correct promoter
site, it:a) dissociates from the core enzyme b) guides the
polymerase through the transcription processc) signals the
polymerase when to terminate transcription and fall off the
DNAd) reassociates with the holoenzyme

A

The end of an RNA from a bacterial gene is likely to containa)
Inverted repeats and a string of U�sb) Direct repeats and a
string of A�sc) Inverted repeats and a string of A�sd)
Direct repeats and a string of U�se) Direct repeats and a string
of G�s

A

Hershey and Chase differentiated between DNA and protein by:A.
labeling the DNA with 32Phosphorous, proteins with 35SulfurB.
labeling the DNA with 35Sulfur, proteins with 32PhosphorousC.
labeling the DNA with cesium , proteins with chlorideD. labeling
the DNA with 14Carbon, proteins with 3Hydrogen

A

Which is a mechanism that allows a single gene to encode more than
one polypeptide?A. Regulation of mRNA stabilityB.
Alternative RNA splicing C. RNA interferenceD. Reverse
transcriptionE. None of the above

B

Is the �replace RNA� enzyme AA DNA Pol IB DNA Pol
IIIC HelicaseD PrimaseE Gyrase

A

12) Relieves torsional stress
A DNA Pol IB DNA Pol IIIC HelicaseD
PrimaseE Gyrase

E

13) Actually synthesizes RNA, not DNA
A DNA Pol IB DNA Pol IIIC
HelicaseD PrimaseE Gyrase

D

14) Has 5�-3� exonuclease activity
A DNA Pol IB DNA Pol IIIC
HelicaseD PrimaseE Gyrase

A

15) Opens the double helix for replication machinery
A DNA Pol IB DNA Pol IIIC
HelicaseD PrimaseE Gyrase

C

16) Synthesizes the majority of the DNA during replication
A DNA Pol IB DNA Pol IIIC
HelicaseD PrimaseE Gyrase

B

After several generations of labelling in 15N media, Meselson and
Stahl switched to 14N media for one generation. The DNA
extractedafter one generation in 14N media was centrifuged in a
cesium gradient. The DNA formed a single band with a density somewhere
betweenDNA grown in 15N media and DNA grown in 14N media. This
data is consistent with:A. conservative replicationB.
semiconservative replicationC. dispersive replicationD.
conservative and dispersive replicationE. semiconservative and
dispersive replication

E

An in vitro transcription system that contains a bacterial gene
initiates transcription, but from random points on the DNA. Which of
thefollowing proteins most likely is missing from the
reaction?A. TATA-binding proteinB. RNA polymerase
IIC. rho factorD. sigma factor

D

The DNA replication enzyme whose function most closely resembles RNA
polymerase isA. DNA Polymerase IB. DNA Polymerase
IIIC. Primase D. TelomeraseE. Helicase

C

For double-stranded DNA, which of the following base ratios usually
does not equal 1?A. (A+T)/(G+C)B. (A+G)/(C+T)C.
C/GD. (G+T)/(A+C)E. A/T

A

An siRNA is processed by which enzyme into the 21 base pair
fragment?A. SpicerB. CutterC. CleaverD. Dicer
E. RISC

D

Given the DNA sequence 3�-ACGCTACGTC-5� which is properly hybridized
to its complementary strand, how many hydrogen bondshold the two
strands together?A. 10B. 16C. 24D. 26E. 30

D

A DNA molecule is 900 bp long and has 100 complete rotations. This
molecule isa. Positively supercoiled b. Negatively
supercoiledc. Relaxed

A

Which of the following need(s) to match correctly an anticodon and an
amino acid?A. CodonB. mRNAC. Aminoacyl-tRNA
synthetase D. RibosomeE. More than one of the above

C

The genetic code is universal except forA. prokaryotes, which
use a different genetic code than eukaryotes.B. a few
mitochondrial genes, which substitute one sense codon for
another.C. viruses, which use an entirely different genetic
code.D. archaebacteria, which have their own genetic code.

B

To translate a mRNA you require two other RNAs. These areA.
tRNA and mRNAB. tRNA and miRNAC. rRNA and tRNA D.
rRNA and snRNA

C

Which of the following ratios of molar quantities of histones in DNA
would be the largest?a) H3/H4b) H2A/H2Bc)
H1/H4d) H2A/H1 e) H4/H3

D

What would Avery, Macleod, and McCarty have concluded if their
results had been that only RNAse treatment of the heat-killed
bacteriaprevented transformation of genetic virulence?a.
that DNA was the genetic materialb. that RNA was the genetic
materialc. that protein was the genetic materiald. that
DNAse or protease, but not RNase, stimulates transformation

B

Suppose that Hershey and Chase found that phage ghosts contained 32P
label but the label was absent from infected E.
coli.Furthermore, suppose they found 35S lacking in the ghosts
and present in the infected E. coli. They would have
concluded:a. that protein was the genetic material in phage.
b. that DNA was the genetic material in phage.c. that
somehow the radioactivity prevented DNA from getting into E.
coli.d. that protein and DNA together made up the genetic material.

A

What linkage is present at the 5� cap of an mRNA?a) 3� to
3�b) 5� to 5� c) 5� to 3�d) 3� to 5�e) 5� to 2�

B

What is one difference between DNA replication in bacteria versus
eukaryotes?A) Eukaryotic chromosomes are replicated
bidirectionally, while bacterial chromosomes are replicated in one
direction.B) Eukaryotic chromosomes have many origins of
replication and replicate bidirectionally, while bacteria have only
one origin of replicationand replicate unidirectionally.C)
Bacterial chromosomes are replicated bidirectionally, while eukaryotic
chromosomes are replicated in one direction.D) Eukaryotic
chromosomes have many origins of replication, while bacteria have only
one origin of replication. E) The process is identical in
bacterial and eukaryotic DNA replication.

D

What enzyme(s) is/are responsible for removal of RNA primers and
joining of Okazaki fragments?A) DNA polymerase IB) DNA
polymerase IIIC) DNA ligaseD) DNA polymerase I and DNA
ligase E) DNA polymerase III and DNA ligase

D

During initiation of DNA replication in E. coli, what is the role of
helicase?a. It binds to the origin and causes a short section of
the double helix to unwind.b. It binds to and stabilizes the
single-stranded DNA.c. It reduces torsional strain by making
double-stranded DNA breaks ahead of the replication fork.d. It
breaks hydrogen bonds between the two DNA strands.

D

In human beings, which sequence will have the largest Cot
1/2?A. telomeresB. parasitic DNAC. ribosomal RNA
geneD. a gene involved in eye color

D

RNA interference blocks further transcription of the target gene
byA. Methylating the histonesB. Methylating the cytosines
in the geneC. Degrading the nucleotides of the geneD.
Binding on to the DNA to block transcription

B

A chromosome with a centromere in the middle is calledA.
metacentric B. submetacentricC. acrocentricD.
telocentricE. centrocentric

A

Which of the following is the indivisible unit of genetic
informationa) gene b) allelec) chromosomed) ribosome

A

A man expressing a Y-linked trait marries a woman who does not
express the trait. What do the rules of probability indicatefor
the ratio of their offspring?a) All sons and all daughters
affectedb) Half of sons and half of daughters affectedc)
Half of sons and all daughters affectedd) All sons and half of
daughters affectede) All sons and no daughters affected

E

In a cross of AaBbCcDdEe X AaBbccDDEe, what proportion will have the
ABCde phenotype?a) 0 b) 1/4c) 27/64d)
27/128e) 37/128

A

B is a dominant giving black color. b is the recessive allele causing
blue color. In a population of 100 B/bindividuals, ten purple
(blue mixed with black) animals are seen. This is an example
ofA. incomplete dominanceB. partial dominanceC.
reduced penetranceD. reduced expressivity E. poor record keeping

D

In a cross of two double heterozygotes A/a B/b, what proportion will
be dominant for a, recessive for b?A. 1/16B. 3/16 C.
4/16D. 5/16E. 9/16

B

The chromatids in a pair of chromosomes are held together at a
specific region of the chromosome called theA. Centromere
B. Mitotic spindleC. CentrosomeD. Chromosome binding site

A

A cross of two heterozygous individuals produces 78 dominants and 22
recessives. What is the chi-square valuefor these
results?A. 8/25B. 16/75C. 4/75D. 8/75E. 12/25

E

Which of the following results from a physical exchange between
chromatids of homologous chromosomes?A. BivalentB. Meiotic
divisionC. Chiasma D. TetradE. Synapsis

C

In a cross of two flies +/vg Cy/+ +/se +/ri X +/vg Cy/+ se/se +/ri
what proportion of the offspring will bemutant in phenotype for
all four markers?A. 0B. 1/4C. 1/32D.
1/64E. 3/128

E

Which of the following occurs only in Meiosis II:A. Separation
of homologuesB. Synaptonemal complex formationC. Formation
of spindle polesD. Division of centromeres E. None of the above

D

Women who are heterozygous for an X-linked condition causing the
absence of sweat glands exhibit patches ofskin with sweat glands
and patches lacking sweat glands. This patchwork effect is the result
ofA. MosaicismB. Mitotic segregationC. All
heterozygotes would be expected to display this phenomenon regardless
of sexD. Dosage compensationE. Both A and D

E

Color blindness is X-linked recessive and blood type is autosomal. If
two parents who are both Type A and havenormal vision produce a
son who is color blind and type O, what is the probability that their
next child will be a sonwho has normal vision and is blood type
A?A. 0B. 1/8C. 1/4D. 3/16 E. 1

D

A drosophila with 3 X chromosomes, a Y chromosome and 2 sets of
autosomes will be aA. normal maleB. normal femaleC.
metafemaleD. intersexE. metamale

C

On average, what proportion of the X-linked genes in a daughter are
the same as her father?a) 0b) 1/4c) 1/2 d)
3/4e) 1

C

XYY can only occur because of a normal gamete fusing with a gamete
which is the result of nondisjunction in:A. Meiosis II in
maleB. Meiosis I and II in maleC. Meiosis I in
maleD. Meiosis II in femaleE. Meiosis I and II in female

A

Which of the following is not a reason that the garden pea was a good
genetic organismA. rapid breeding rateB. true-breeding
linesC. easily identified traitsD. multiple
chromosomesE. ability to control mating

D

white (w) is an X-linked recessive and tinman (tn) is an autosomal
recessive mutation. What proportion of white,tinman females
(relative to whole population) is expected in the F2 starting with a
true breeding white female which iswild type for tinman mating
with a true breeding male mutant only for tinman.a) 0b)
1/32c) 1/16d) 1/8e) 3/16

C

A drosophila with 2 X chromosomes, a Y chromosome and 3 sets of
autosomes will be aA. normal maleB. normal femaleC.
metafemaleD. intersex E. metamale

D

The mitotic stage in which the chromosomes condense, nuclei
disappear, and the mitotic spindle forms is knownasA.
MetaphaseB. Prophase C. AnaphaseD. Cytokinesis

B

A colorblind mother and normal vision father have a child who is
XXXXYY and colorblind. This is the result of:A. Nondisjunction
in Meiosis I in the fatherB. Nondisjunction in Meiosis I in the
father and motherC. Nondisjunction in Meiosis I in the father
and Meiosis I and II in the motherD. Nondisjunction in Meiosis
II in the father and Meiosis I and II in the mother. E.
Nondisjunction in Meiosis I and II in the mother

D

A husband and wife have normal vision, although both of their fathers
are colorblind. What is theprobability that their first child
will be a daughter who is colorblind?A. 0B. 1/8C.
1/4D. 1/2E. 1

A

A cross of two heterozygous individuals produces 77 dominants and 23
recessives. What is the chisquarevalue for these
results?A. 8/25B. 16/75 C. 4/75D. 8/75E. 16/25

B

The situation in which an expected phenotype is not expressed in the
numbers predicted is referred toasA. PleiotropyB.
Variable expressivityC. Incomplete penetrance D. Recessivity

C

A colorblind mother and normal vision father have a child who is
XXXXYY and colorblind. This is theresult of:A.
Nondisjunction in Meiosis I in the fatherB. Nondisjunction in
Meiosis I in the father and motherC. Nondisjunction in Meiosis I
in the father and Meiosis I and II in the motherD.
Nondisjunction in Meiosis II in the father and Meiosis I and II in the
mother. E. Nondisjunction in Meiosis I and II in the mother

D

Mating of two organisms produces a 1:1 ratio of phenotypes in the
progeny. The parental genotypes areA. Aa x AaB. Aa x aa
C. AA x aaD. AA x AA

B

Color blindness is X-linked recessive and blood type is autosomal. If
two parents who are both Type Aand have normal vision produce a
son who is color blind and type O, what is the probability that their
nextchild will be a female who has normal vision and is type
O?A. 0B. 1/8C. 1/4D. 1/2E. 1

B

Two parents with normal vision have a child with only one sex
chromosome. The child is colorblind.This is the result of
nondisjuntion in:A. Either Meiosis I or II in maleB.
Meiosis II in maleC. Either Meiosis I or II in femaleD.
Meiosis I in femaleE. Meiosis I in male

A

36) Metaphase A. Chromosome replication occursB. Sister
chromatids begin to separateC. Chromosomes become
alignedD. Replicated chromosomes begin to condenseE.
Cytoplasm divides

C

37) Anaphase A. Chromosome replication occursB. Sister
chromatids begin to separateC. Chromosomes become
alignedD. Replicated chromosomes begin to condenseE.
Cytoplasm divides

B

38) Telophase EA. Chromosome replication occursB. Sister
chromatids begin to separateC. Chromosomes become
alignedD. Replicated chromosomes begin to condenseE.
Cytoplasm divides

E

39) Interphase A. Chromosome replication occursB. Sister
chromatids begin to separateC. Chromosomes become
alignedD. Replicated chromosomes begin to condenseE.
Cytoplasm divides

A

40) Prophase
A. Chromosome replication occursB. Sister
chromatids begin to separateC. Chromosomes become
alignedD. Replicated chromosomes begin to condenseE.
Cytoplasm divides

D

A chromosome with a centromere close to the end is calledA.
metacentricB. submetacentricC. acrocentric D.
telocentricE. centrocentric

C

Which of the following can result in dicentric chromosomes during
meiosis?A. duplicationB. paracentric inversionC.
pericentric inversionD. deletionE. translocation

B

Mendel's second law or fourth postulate states that alleles of
different genes segregate independently of eachother. At the
chromosome level this is because:a) Synthesis of chromosomes in
Interphaseb) Chiasma forming during Prophase Ic)
Chromosomes aligning in metaphase Id) Chromosomes aligning in
metaphase IIe) chromosomes separating in anaphase II

C

Agouti yellow and Siamese coat color are examples ofA.
interdominanceB. codominanceC. pleitropyD.
incomplete dominanceE. lethal genes

C

The following progeny phenotypes result from a cross of an A/a B/b
individual with an a/a b/b individualA B 400a b 400A
b 100a B 100What is the recombination rate between A and
B?A. 0.1B. 0.2 C. 0.3D. 0.4

B

In mammalian cells, the only autosomal chromosome abnormality found
at any frequency isA) monosomyB) trisomy C)
autopolyploidyD) allopolyploid

B

True breeding white flowering plants are crossed to true breeding red
flowering plants. The offspring are red flowered.When these
flowers are self crossed, the resultant offspring are 147 red and 12
white flowered plants. Which of the followingpathways is most
likely the cause of thisA. white
---A--->pink---B--->redB. pink
---A--->pink---B--->redC. red<---A---white
---B--->redD. white ---A--->pink white---B--->pink both
genes producing pink together give red

C

Gene A converts compound A (green) to compound B (blue). Compound A
is lethal to the organism if not broken downthrough this
process. Gene B converts compound B to compound C (red). These
compounds determine the color of theindividual. In a self cross
of A/a B/b individuals, what proportion of green to blue offspring
should be seen?A. 1:16B. 3:1C. 9:3:3:1D. 0:4
E. 4:3

D

Which of the following is the most likely description of a trait
which shows up rarely in a family tree equally in both sexesand
only after a consanguineous mating?A. Recessive autosomal
B. Dominant AutosomalC. X-LinkedD. Y-linkedE.
Either X or Y-linked

A

An individual has four X chromosomes and no Y but appears to be a
male. This is likely due to:A. Love of remote controlsB.
Transferred testosterone receptorC. Transferred PARD.
Transferred SRY

D

In a population of 10,000 individuals, 200 men are afflicted with a
recessive, X-linked disease. How many woman would beexpected to
be afflicted in this population?A. 1B. 2C. 4D.
8 E. 24

D

Disease A affects .0004 of people, Disease B affects .0009 of people,
X-linked Disease C affects .0025 of males. The allelefrequencies
for the recessive, disease causing alleles are:A. .0004 for a,
.0009 for b, .0025 for cB. .02 for a, .03 for b, .05 for
cC. .04 for a, .02 for b, .0025 for cD. .02 for a, .03 for
b, .0025 for cE. .03 for a, .02 for b, .05 for c

D

Joe the Farmer decides to increase the weight of the eggs produced by
his chickens. He knows that the heritability of eggweight is
0.6. If he selects for chickens producing eggs which are 5 grams
heavier than the average, how much heavier will bethe eggs
produced by their offspring?A. 1B. 2C. 3
D. 4E. 5

C

True breeding white flowering plants are crossed to true breeding red
flowering plants. The offspring are redflowered. When these
flowers are self crossed, the resultant offspring are 87 red and 72
pink flowered plants. Whichof the following pathways is most
likely the cause of thisA. white
---A--->pink---B--->redB. pink
---A--->pink---B--->red C. red<---A---pink
---B--->redD. white ---A--->pink white---B--->pink both
genes producing pink together give red

B

In a population at Hardy-Weinberg equilibrium, 25% of the individuals
have the recessive phenotype. What is the percentageof
heterozygotes in the population?a. 24%b. 36%c.
48%d. 50% e. 72%

D

n a population of 10,000 individuals, 100 express a recessive trait.
How many homozygotes for the dominant allele are thereexpected
to be?A. 600
B. 1,000C. 1,800D. 6,400E. 8,100

E

11) Maintains the single-strand form of DNA A SSBB DNA
Pol IIIC HelicaseD PrimaseE Telomerase

A

12) Is often overactive in cancer cells A SSBB DNA Pol
IIIC HelicaseD PrimaseE Telomerase

E

4) Coordinate replication requires two A SSBB DNA Pol
IIIC HelicaseD PrimaseE Telomerase

B

15) Opens the double helix for replication machinery A SSBB DNA
Pol IIIC HelicaseD PrimaseE Telomerase

C

XXXXY can only occur because of a normal gamete fusing with a gamete
which is the result of nondisjunction in:A. Meiosis II in
maleB. Meiosis I and II in maleC. Meiosis I in
femaleD. Meiosis II in femaleE. Meiosis I and II in female

E

In the Meselson Stahl experiment differentiating the modes of
replication by labeling with different forms of nitrogen and spinning
in acesium gradient, how many rounds of replication were needed
to determine if the following modes of replication were not used in E.
coli?A. one round of replication for dispersive; one round of
replication for conservativeB. one round of replication for
dispersive; two rounds of replication for conservativeC. two
rounds of replication for dispersive; one round of replication for
conservativeD. two rounds of replication for dispersive; two
rounds of replication for conservative

C

In what form of bacterial recombination is a double recombination
event not required to integrate the DNA into the chromosome?A.
ConjugationB. TransformationC. TransductionD.
Required in all formsE. Not required in any forms

D

The form of DNA normally found in our cells is:A. AB. B
C. CD. DE. E

B

The function of DNA ligase is toA. Catalyze the formation of
hydrogen bonds between adjacent 5'-P and 3'-OH terminiB. Relax
supercoiling of DNAC. Add methyl groups to DNAD.
Facilitate base pairing between single-stranded molecules of
DNAE. Catalyze the formation of covalent bonds between adjacent
5'-P and 3'-OH termini

E

The pyrimidine bases areA. Thymine and cytosineB. Thymine
and guanineC. Adenine and guanineD. Cytosine and
adenineE. Cytosine and guanine

A

Acts in a distance and orientation independent fashion a)CAAT
box or GC boxb) enhancerc) TF IIDd) RNA Polymerase
Ie) RNA Polymerase III

B

32) Produces large quantities of a few large polymers CAAT box
or GC boxb) enhancerc) TF IIDd) RNA Polymerase
Ie) RNA Polymerase III

D

33) upstream elements CAAT box or GC boxb)
enhancerc) TF IIDd) RNA Polymerase Ie) RNA
Polymerase III

A

34) Makes tRNA CAAT box or GC boxb) enhancerc) TF
IIDd) RNA Polymerase Ie) RNA Polymerase III

E

35) a protein which binds to TATA boxes
a)CAAT box or GC boxb) enhancerc) TF
IIDd) RNA Polymerase Ie) RNA Polymerase III

C

produces mRNA A) middle repetitive DNAB) unique
DNAC) A siteD) EF-TuE) RF I

B

38) a protein shaped like an RNA A) middle repetitive
DNAB) unique DNAC) A siteD) EF-TuE) RF I

E

39) Required for translocation of ribosome A) middle repetitive
DNAB) unique DNAC) A siteD) EF-TuE) RF I

D

40) tRNA entry site A) middle repetitive DNAB) unique
DNAC) A siteD) EF-TuE) RF II

C

41) ribosomal RNA genes
A) middle repetitive DNAB) unique DNAC) A
siteD) EF-TuE) RF Imiddle repetitive DNA

A

4) needed for splicing a) CAAT box or GC boxb)
snRNPc) TF IIDd) 5� cape) telomerase

B

5) contains a triphosphate phosphodiester linkage a) CAAT box
or GC boxb) snRNPc) TF IIDd) 5� cape) telomerase

D

6) upstream elements a) CAAT box or GC boxb)
snRNPc) TF IIDd) 5� cape) telomerase

A

7) needed for translation a) CAAT box or GC boxb)
snRNPc) TF IIDd) 5� cape) telomerase

D

8) a protein which binds to TATA boxes a) CAAT box or GC
boxb) snRNPc) TF IIDd) 5� cape) telomerase

C

9) contains an RNA which is used as a template
a) CAAT box or GC boxb) snRNPc) TF
IIDd) 5� cape) telomerase

E

The F factor is a:a) genomeb) plasmid c)
plaqued) prophagee) lawn

B

Which of the following is not an RNA involved in transcription or
translation?a) tRNAb) mRNAc) gRNAd)
hRNAe) snRNA

D

The genetic code is said to be degenerate becauseA. mRNA is
rapidly degradedB. The code is not universal among
organismsC. Several codons direct the insertion of the same
amino acid into a polypeptide chain D. Frameshift mutations are
toleratedE. Stop codons may have corresponding tRNA molecules

C

Which level of protein structure deals with just the sequence of
amino acids?a) primary b) secondaryc)
tertiaryd) quaternary

A

The P site in the ribosome is named forA. P-Diddy B. Protein C.
Peptidyl D.Polysome

C

Which of the following describes the efficiency of an F� strain in
causing:F- => F+ transfer of geneticinformationA)
High HighB) High LowC) Low HighD) Low Low

A

The signal for polyadenylation of the mRNA isa) AAAAAAb)
AAUUUUc) AAUAAA d) AAAUAAe) AUAUAU

C

27) Participates in addition of an amino acid to a growing
polypeptide chain
a) aminoacyl synthetaseb) EF-Tuc)
tRNAd) RFIIe) IF2

B

28) Interacts with A site when ribosome is on a stop codon a)
aminoacyl synthetaseb) EF-Tuc) tRNAd) RFIIe) IF2

D

29) helps assemble large and small ribosomal subunits a)
aminoacyl synthetaseb) EF-Tuc) tRNAd) RFIIe) IF2

E

30) carries the anticodon a) aminoacyl synthetaseb)
EF-Tuc) tRNAd) RFIIe) IF2

C

31) charges the tRNA a) aminoacyl synthetaseb) EF-Tuc)
tRNAd) RFIIe) IF2

A

The A site in the ribosome is named forA. Amino end B. Attila
the Hun C. Aminoacyl D. Amino Acid

C

The snRNP which binds the 5� splice junction isa) U1b)
U2c) U3d) U4e) U5

A

Discontinuous replicationA. Occurs on the DNA strand
synthesized overall in the 3' to 5' directionB. Occurs on the
DNA strand synthesized overall in the 5' to 3' directionC.
Results in the formation of Okazaki fragmentsD. Both A and
CE. Both B and C

D

Which level of protein structure includes alpha helix and beta
pleated sheet?a) primaryb) secondaryc)
tertiaryd) quaternary

B

Which of the following is not different between eukaryotic and
prokaryotic transcription?a) Splicingb) 5' cappingc)
poly-adenylationd) terminatione) all are different

E

The correct order of activity in replication isa) helicase,
SSB, Pol III, primase, Pol I, ligaseb) primase, helicase, SSB,
Pol III, Pol I, ligasec) SSB, helicase, primase, Pol III, Pol I,
ligased) helicase, SSB, primase, Pol I, Pol III, ligasee)
helicase, SSB, primase, Pol III, Pol I, ligase

E

Which one of the following is not a stop codon?A. UAA B. UGG C.
UAG D. UGA

B

A DNA molecule of 500 bp will have approximately how many
turns?A)50 B)500C)5,000D)250E)25

A

Which of the following statements is true?a) Pol II promoters
and Pol I promoters are internalb) Pol II promoters and Pol III
promoters are internalc) Pol I promoters are at the 5� end and
Pol III promoters are internald) Pol III promoters are at the 5�
end and Pol I promoters are internal

D

Synthetic mRNAs were used to predict the genetic code. Which of the
following is not true?a) Repeats of the same nucleotide produce
a single amino acid polymerb) repeats of two nucleotides produce
a polymer with a two amino acid repeat unitc) repeats of three
nucleotides produce a polymer with a three amino acid repeat unit
d) repeats of four nucleotides produce a polymer with a four
amino acid repeat unit

C

Which level of protein structure deals with interactions of
subunits?a) primaryb) secondaryc) tertiaryd) quaternary

D

Mutant bacteria strains called auxotrophs can only grow in _______
medium.a. completeb. minimalc. partially
supplementedd. no

A

The following figure shows the results of interrupted-mating
experiments with 3 different Hfr strains. What is theorder of
the genes, starting with C?Hfr strain Order of transfer1
A, B, E, D, F2 D, F, C, G, A3 D, E, B, A, Ga. C, G,
A, D, F, B, Eb. C, F, D, B, A, E, Gc. C, B, E, D, F, G,
Ad. C, G, A, B, E, D, F e. C, D, F, G, A, B, E

D

Most cases of Down syndrome arise from:a. inversions.b.
deletions.c. X-rays.d. maternal nondisjuction.e.
unequal crossing over.

D

The most common aneuploidy seen in living humans has to do with
___________.a. autosomesb. sex chromosomesc.
chromosome 21d. chromosome 13e. None of the above

B

Which of these RNA sequences could form a hairpin also called a stem
loop?a. 5? GGGGTTTTCCCC 3? b. 5? AAAAAAAAAAAA 3?c.
5? ACACACACACAC 3?d. 5? TTTTTTCCCCCC 3?

A

Which of the following describes the efficiency of an Hfr strain in
causing:F- => F+ transfer of geneticinformationA)
High HighB) High LowC) Low High D) Low Low

C

An in vitro transcription system transcribes a bacterial gene but
terminates inefficiently. What is one possibleproblem?a.
There is a mutation in the �10 consensus sequence, which is required
for efficient termination.b. Rho factor has not been added.
c. Sigma factor has not been added.d. Spliceosomes have
not been added

B

If the sequence of an RNA molecule is 5�-GGCAUCGACG-3�, what is the
sequence of the template strand ofDNA?a.
5�-GGCATCGACG-3�b. 3�-GGCATCGACG-5�c.
5�-CCGTAGCTGC-3�d. 5�-CGTCGATGCC-3� e. None of the above

D

Which is a mechanism that allows a single gene to encode more than
one polypeptide?a. Regulation of mRNA stabilityb.
Alternative RNA splicing c. RNA interferenced. Reverse
transcriptione. None of the above

B

25. Okazaki fragmentsa. supercoil removalb. RNA primer
synthesisc. 3? -->5? exonuclease activityd. lagging
strande. leading strand

D

26. DNA primase
a. supercoil removalb. RNA primer
synthesisc. 3? -->5? exonuclease activityd. lagging
strande. leading strand

B

27. DNA gyrase a. supercoil removalb. RNA primer
synthesisc. 3? -->5? exonuclease activityd. lagging
strande. leading strand

A

28. continuous synthesis
a. supercoil removalb. RNA primer
synthesisc. 3? -->5? exonuclease activityd. lagging
strande. leading strand

E

29. DNA proofreading
a. supercoil removalb. RNA primer
synthesisc. 3? -->5? exonuclease activityd. lagging
strande. leading strand

C

30. heteroduplex DNAa 5� -> 3� polymerase activityb.
strand invasionc. phosphodiester bonds at DNA nicksd.
5�->3� exonuclease activitye. bidirectional circular replication

B

31. DNA ligase a 5� -> 3� polymerase activityb. strand
invasionc. phosphodiester bonds at DNA nicksd. 5�->3�
exonuclease activitye. bidirectional circular replication

C

32. theta replication a 5� -> 3� polymerase activityb.
strand invasionc. phosphodiester bonds at DNA nicksd.
5�->3� exonuclease activitye. bidirectional circular replication

E

33. DNA synthesisa 5� -> 3� polymerase activityb.
strand invasionc. phosphodiester bonds at DNA nicksd.
5�->3� exonuclease activitye. bidirectional circular replication

A

34. Primer removal
a 5� -> 3� polymerase activityb. strand
invasionc. phosphodiester bonds at DNA nicksd. 5�->3�
exonuclease activitye. bidirectional circular replication

D

5) Actually synthesizes RNA, not DNA A DNA Pol IIB
TelomeraseC HelicaseD PrimaseE Gyrase

D

6) Maintains ends of chromosomes A DNA Pol IIB
TelomeraseC HelicaseD PrimaseE Gyrase

B

7) Opens the double helix for replication machinery A DNA Pol
IIB TelomeraseC HelicaseD PrimaseE Gyrase

C

8) Has no required function in DNA chromosome replicationA DNA
Pol IIB TelomeraseC HelicaseD PrimaseE Gyrase

A

9) contains an RNA molecule which is required for function
A DNA Pol IIB TelomeraseC HelicaseD
PrimaseE Gyrase

B

10) Relieves torsional stress
A DNA Pol IIB TelomeraseC HelicaseD
PrimaseE Gyrase

E

Termination of transcription in prokaryotes is carried out in part by
aA) trinucleotide repeatB. LINE repeatC. stem loop
or hairpin structure D. Bruce Bowen

C

23) Which of the following statements regarding gene expression is
true?A. Messenger RNA is translated in the 5' to 3' direction
B. Transcription involves the association of mRNA with
ribosomesC. mRNA undergoes proof-readingD. Prokaryotic RNA
usually undergoes nuclear processingE. Polypeptides are
synthesized by addition of amino acids to the amino terminus

A

The genetic code is said to be redundant becauseA. mRNA is
rapidly degradedB. The code is not universal among
organismsC. Several codons direct the insertion of the same
amino acid into a polypeptide chain D. Frameshift mutations are
toleratedE. Stop codons may have corresponding tRNA molecules

C

Which of the following is not present as a dimer in a histone
core?A. H1B. H2AC. H2BD. H3E. H4

A

A DNA molecule of 500 bp will have approximately how many
turns?A)50 B)500C)5,000D)250E)25

A

A drosophila with 4 X chromosomes, a Y chromosome and 4 sets of
autosomes will be aA. normal maleB. normal femaleC.
metafemaleD. intersexE. metamale

B

Bacterial cells containing an F plasmid that has acquired bacterial
chromosomal genes are called:a. F+.b. F?. c.
F�.d. Hfr.

B

Translation is ended because of the action ofA. Initiation
factorsB. Elongation factorsC. Termination factorsD.
Release factorsE. Dissociation factors

D

In watchamakallits black fur is dominant to white. In a certain
population of watchamakallits, white fur occurs in a frequency
of1/100. What is the frequency of the recessive allele in the
population, assuming Hardy-Weinberg equilibrium?a. 0.01b.
0.1 c. 0.5d. 0.9e. 0.9947. What is the
frequency of heterozygous watchamakallits in the population described
in question 46, assuming Hardy-Weinbergequilibrium?a.
0.01b. 0.02c. 0.18 d. 0.82e. 0.99

B,C

bacterial cell transfers chromosomal genes to F� cells, but it rarely
causes them to become F+. The bacterial cell is:a. Hfr.b.
lysogenic.c. auxtrophic.d. lytic.

A

Which agent of evolution would tend to increase the frequency of all
recessive phenotypes in a population relative to what isexpected
from Hardy-Weinberg equilibrium?a. Genetic driftb.
Inbreeding c. Natural selectiond. Positive assortative
matinge. Negative assortative mating

B

A chromosome with a centromere near but not at the end is
calledA. metacentricB. submetacentricC. acrocentric
D. telocentricE. centrocentric

C

Pick the statement that is false.a. A gene is the fundamental
unit of heredity.b. A genome is the complete set of DNA carried
in an organism.c. Alleles are alternative forms of a
gene.d. Genotype refers to the genes of an organism that
determine a particular trait.e. Gene pool is the total of all
genes in an individual.

E

If adenine makes up 15 percent of the bases in a specific DNA
molecule, what percentage of the bases are cytosine.a.
15%b. 30%c. 35% d. 60%e. 70%

C

Assume that a cross is made between AaBb and aabb plants and all of
the offspring areeitherAaBb or aabb. These results are
consistent with the following circumstance:A. Complete
linkageB. Alternation of generationsC. CodominanceD.
Incomplete dominanceE. Hemizygosity

A

DNA polymerase requires a 3� OH group provided by a primer before it
can begin DNA synthesis. How is it that the primer can be synthesized
on the DNA without a 3� OH group?

The primer is RNA-synthesized by RNA polymerase, which does not
require a 3� OH group.

Mismatch repair requires the ability to distinguish between template
and newly synthesized DNA strands. How can E. colidistinguish
between these two strands?

Template DNA is methylated

Eukaryotic DNA replication is similar to prokaryotic DNA replication.
However, eukaryotes require unique replication strategies because of
unique features of eukaryotic DNA. Which is NOT a unique feature
linked to replication strategies?

Eukaryotic DNA contains introns.

Why is it that low-fidelity eukaryotic DNA polymerases are able to
replicate DNA that contains abnormal bases, distorted structures, or
bulky lesions, whereas high-fidelity DNA polymerases stall at these areas?

Low-fidelity DNA polymerases have large active sites that can bind to
these regions.

You identify a mutant whose chromosomes shorten after each round of
replication. A mutation in which gene would explain this observation?

telomerase

Understanding DNA recombination is important in understanding ___________.
a. DNA repair
b. gene linkage traits
c. genetic variation
d. all of the above

D

How would DNA replication be affected in a cell that is lacking topoisomerase?

Torsional strain ahead of the replication fork would eventually cause
replication to stop

Which of the following is true of RNA compared to DNA?

RNA has a hydroxyl group on the 2�-carbon atom of its sugar component....

When RNA is transcribed from a gene, which strand of DNA is used?

the template strand

The transcription unit includes three essential regions. What is the
proper order of these regions?

promoter, RNA coding sequence, terminator

What would the result be if a specific sigma subunit were mutated?

RNA polymerase would fail to initiate transcription at the promoter
specific to the sigma subunit

The bacterial holoenzyme binds to which part of the promoter?

�10 and �35 consensus sequence

In rho-dependent transcription termination, the rho factor binds to ___________.

mRNA

Which is NOT one of the DNA sequences known to regulate gene transcription?

basal transcription apparatus

In eukaryotes, what initially binds to the TATA box on the DNA template?

TFIID

RNA polymerase I, II, and III terminate transcription differently.
Which of the following statements is most correct?
a. RNA polymerase I requires a termination factor (like rho factor).
b. RNA polymerase II synthesizes hundreds of bases beyond
what is needed for the mRNA, which is degraded by the protein Rat1.
c. RNA polymerase III terminates transcription after a
termination sequence is transcribed.
d. All of the above statements are correct.......

D.

What is the most likely result from mutating a prokaryotic
Shine�Dalgarno sequence?

The ribosome would not be able to bind to the mRNA.

Which mRNA processing event adds stability to the mRNA?

-addition of a 5� cap -addition of a poly(A) tail to the 3� end

What kind of RNA functions in splicing and is associated with the spliceosome?

small nuclear RNA (snRNA)

Which intron component is the first to be cleaved during the splicing process?

5� splice site

What is it called when mRNA splicing occurs that results in variously
sized mRNAs?

alternative 3� cleavage sites

What process causes a protein amino acid sequence NOT to correspond
with its DNA sequence?

RNA editing

Which arm of tRNA binds to mRNA?

anticodon arm

Which is NOT true of ribosomes?

They are made up of only ribosomal RNA molecules.

Small interfering RNA (siRNA) molecules block gene expression by
degrading mRNA or inhibiting transcription. Which genes do they target?

genes from which they were transcribed

a. 5' untranslated region

ribosome binding site or Shine-Dalgarno sequence is found within the
5'untranslatedregion. However, eukaryotic mRNA does not have the
equivalent sequence, and aeukaryotic ribosome binds at the 5' cap of
the mRNA molecule.

b. Promoter

b. The promoter is the DNA sequence that the transcription apparatus
recognizes and binds to initiate transcription.

c. AAUAAA consensus sequence

c. The AAUAAA consensus sequence lies downstream of the coding region
of the gene. It determines the location of the 3' cleavage site in the
pre-mRNA molecule.

d. Transcription start site

d. The transcription start site begins the coding region of the gene
and is located 25 to 30 nucleotides downstream of the TATA box.

e. 3' untranslated region

e. The 3' untranslated region is a sequence of nucleotides at the 3'
end of the mRNA that is not translated into proteins. However, it does
affect the translation of the mRNA molecule as well as the stability
of the mRNA.

f. Introns

f. Introns are noncoding sequences of DNA that intervene within
coding regions of a gene.

h. Poly(A) tail

h. A poly(A) tail is added to the 3' end of the pre-mRNA. It affects
mRNA stability.

i. 5' cap

i. The 5' cap functions in the initiation of translation and mRNA stability.

What kind of chemical bond holds adjacent amino acids together?

a peptide bond

The wobble rules account for non�Watson and Crick nucleotide pairing
at which position of the codon and anticodon, respectively?

third and first

Any given DNA sequence has ______ possible reading frames, and the
correct one is set by a(n) ___________________.

3; initiation codon

What is the final component of the initiation complex to be added?

a large subunit of the ribosome

The formation of peptide bonds during elongation is catalyzed by an
enzyme called the ribozyme. The ribozyme is made of what macromolecule?

rRNA in the large subunit of the ribosome

Simultaneous transcription and translation does NOT occur in which of
the following locations?

cytoplasm of eukaryotes

What are isoaccepting tRNAs?

Isoaccepting tRNAs are tRNA molecules that have different anticodon
sequences but accept the same amino acids.

Which of the following chromatin changes will repress gene
transcription?

DNA not methylated

which of the following is true of transcriptional activator proteins?

Some have acetyltransferase activity

What is the role of insulators?

They block the role of enhancers.

Which two mRNA structures interact with each other to provide mRNA stability?

5� cap and 3� poly(A) tail

siRNA and miRNA inhibit gene expression by all of the following EXCEPT

stabilizing translation machinery

When present in small amounts in sequencing reactions,
dideoxyribonucleoside triphosphates (ddNTPs) terminate the sequencing
reaction at different positions in the growing DNA strands. ddNTPs
stop a sequencing reaction because they:

c. lack a hydroxyl (�OH) group at their 3? end

All of the following are true of single nucleotide polymorphisms
(SNPs) EXCEPT

they are rare among persons of the same ethnic group.

Shotgun sequencing does NOT:

depend on physical or genetic maps

Genetic and physical maps differ
a. in the order of the genes.
b. in that physical maps are based on recombination frequencies.
c. in that genetic maps are based on base pairs.
d None of the above.

D

Epigenetics is the study of:

inheritance of traits not coded by DNA sequence

How would you expect DNA methylation to alter gene expression

Measurably decrease expression

Cancer is a heterogeneous group of disorders resulting from
a. rapid uncontrolled cell division.
b. a multistep process that requires several different mutations.
c. Both a and b. :

C

Considering the clonal evolution of tumors, mutations in which kinds
of genes would speed up the rate of accumulation of additional mutations?

DNA-repair genes

Which of the following is true of tumor-suppressor genes?
Their normal function is to promote cell proliferation.
b. Their mutant forms are typically dominant
c. They were the first cancer-causing genes to be identified.
d. None of the above is true of tumor-suppressor genes

D

In an experiment you mutate the retinoblastoma gene (RB) such that
its gene product behaves as if hyperphosphorylated. The result would
be a ______ association between RB and E2F, with _______ transcription
of genes necessary for DNA replication.

weaker; continuous stimulation of

Which of the following is the major event associated with the
retinoblastoma cancer?

Both copies of a tumor-suppressor gene are inactivated

Which of the following result(s) directly from metastasis?

Secondary tumors

Which of the following statements is false?

Most tumors arise from germ-line mutations that accumulate during our
life span

Normal cellular genes whose products are involved in facilitating
cell division to occur under appropriate conditions are called?

Proto-oncogenes

Most body (nonreproductive) cells of humans and other multicellular
eukaryotes have two sets of each chromosome. Such cells are ______ and
the matching pairs of chromosomes are called _________.

a. diploid; homologous chromosomes

3. Meiosis provides genetic variability by:
a. crossing over during prophase I.
b. random assortment of homologs during anaphase I.
C. Both a and b

C

Are mutations that change chromosome numbers and structure a cause of
cancer or the result of cancer
a. cause
b. result
c. both

C

According to Mendel�s second law, when the different alleles for one
trait separate into gametes, their separation:

is independent of how different alleles for other traits separate

When Morgan crossed a red-eyed female with a white-eyed male, which
results made Morgan think that the locus affecting eye color was on
the X chromosome?

In the F2, all females had red eyes, and half of the males had red
eyes and the other half had white eyes.

he modification of the amount of protein produced by a sex chromosome
is called:

dosage compensation.

The difference between dominance and epistasis is that:

epistasis masks genes at different loci.

In a certain species of plant, flowers occur in three colors: blue,
pink, and white. A pure-breeding pink plant is mated with a
pure-breeding white plant. All of the F1 are blue. When the blue F1
plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white.
What is the name for this type of interaction?

Recessive epistasis

A mother with blood type AB has a child with blood type B. Give all
possible blood types for the father of this child

A, B, AB, O

You are studying cystic fibrosis (CF). While looking at a pedigree
you notice that the CF phenotype is not present in a set of parents,
but one out of their five children has CF. What can you conclude about CF?

The trait is autosomal recessive.

If genes A and B are linked, what is the maximum percentage of
recombinant gametes that can be produced if a single crossover occurs
during gametogenesis?

50%

What is the most likely order of the linked genes R, S, and T if the
distance between R and S is 22 m.u., the distance between S and T is 8
m.u., and the distance between R and T is 14 m.u.?

S T R

A testcross include

one parent who shows the dominant phenotype for one or more genes and
a second parent who is homozygous recessive for these genes.

Which is NOT a type of chromosomal mutation?

point mutation

After the first round of replication, Meselson and Stahl saw only one
DNA band of density intermediate to DNA containing only 15N or 14N.
After this observation, which hypothesis for DNA replication could be eliminated?

conservative

When RNA is transcribed from a gene, which strand of DNA is used?

the template strand

When a eukaryotic mRNA is hybridized to the complementary DNA, loops
of unhybridized DNA are seen. These loops:
a. correspond to noncoding regions of the gene.
c. demonstrate that genes and proteins are not colinear.
d. Both a and c.

D

ou are studying a biochemical pathway and isolate mutants I, II, and
III. Mutant I can grow if you supplement the medium with Z. Mutant II
can grow if you supplement the medium with X, Y, or Z. Mutant III can
grow if you supplement the medium with X and Z, but not if you
supplement the medium with Y. What is the order of X, Y, and Z in this
biochemical pathway?

Y, X, Z

A mutation in gene X overrides the effect of a previous mutation also
in gene X and restores wild-type phenotype. The second mutation in
gene X is called

intragenic suppressor mutation.

You want to design a repressor protein mutant. Which protein domain
is the best target for preventing binding of
thecorepressor?A. A) DNA-binding domainB. B)
allosteric domain C. C) promoter domainD. D)
helix-turn-helix domainE. E) activator binding site

B

Given the DNA sequence 5?-TAC AAA ATA CAG CGG-3?, which of these
sequences represents a frameshift mutation?A) 5?-TAG AAA ATA CAG
CGG-3?B) 5?-TAC AAA TAC AGC GGG-3? C) 5?-TAC AAG ATA CAG
CGG-3?D) 5?-TAC AAA ATA CAC CGG-3?E) 5?-TAC AAA ATA CAG AGG

B

Which position of a codon evolves at the highest rate?a. All
positions of a codon evolve at the same rate.b. first
positionc. second positiond. third position e. The
first and third positions evolve at the same rate.

D

A mutant E. coli strain, grown under conditions that normally induce
the lac operon, does not produce �-galactosidase.What is a
possible genotype of the cells?a. lacI+ lacP+ lacO+ lacZ+ lacY�
lacA+ b. lacI+ lacP+ lacOc lacZ+ lacY+ lacA+c. lacl+ lacP+
lacO+ lacZ+ lacY+ lacA+d. lacI- lacP+ lacO+ lacZ+ lacY+ lacA+

A

Which type of mutation converts a nucleotide to an alternative
structure with the same composition but slightly different
placementof hydrogen bonds with a rare, less stable form that
causes base-pair mismatch?A) depurinationB)
deaminationC) tautomeric shift #####D) transitionE) transversion

C

You have conducted an Ames test on a given compound. Which of the
following would be classified as a positive result on the
Amestest?A) His- strain grows on an his- plate. B)
His- strain grows on an his+ plate.C) His+ strain grows on an
his- plate.D) His+ strain grows on an his+ plate.E) His+
strain grows on either an his- or an his+ plate.

A

Which of these haploid strains produce ?-galactosidase constitutively
but do not produce permease?A) I- P+ O+ Z+ Y+B) I+ P+ O+
Z- YC)I- P+ O+ Z-Y+D) I+ P+ O- Z+ Y+E) I- P+ O+ Z+ Y-

E

During the attenuation of the trp operon, which stem loop leads to
polycistronic mRNA synthesis during tryptophan starvation?A) 1-3
(antitermination) stem loopB) 3-4 (termination) stem
loopC) 1-2 (pause) stem loopD) 2-3 (antitermination) stem
loopE) 2-4 (termination) stem loop

D

In sporadic cases of retinoblastoma, how many gene mutations are
thought to be necessary in the Rb genes of the samecell for a
tumor to develop?A) oneB) twoC) fourD)
sixE) There is insufficient information to answer this question

B

Nutritional mutations can be defined as ________.A) those
mutations that do not allow a bacterium or fungus to grow on minimal
medium but do allow growth on completemedium B) those
mutations that change the composition of the mediumC) those
mutations belonging to the group called prototrophsD) those
mutations caused by site-specific mutagenesisE) all strains that
are not auxotrophic

A