DAT EXAM REVIEW SOLVED QUESTIONS COMPLETE ANSWERS GRADED A PLUS
Question:
Mark's car began a journey from a point at 7 AM, traveling at 30 mph. At 8 AM Jason's car started
traveling from the same point at 40 mph in the same direction as Jason. At what time will Jason pass
Mark? A. 11 AM B. 10 AM C. 10:30 AM D. 11:30 AM E. Jason never passes Mark.
Answer:
A. 11 AM Let x be the number of hours it takes Jason to pass Mark. During this time, Jason travels
40x miles. Mark started an hour earlier, so he has driven for x+1 hours, during which he travels
30(x+1) miles. Therefore, 40x = 30(x+1) x=3 3 hours from 8 AM is: 11 AM.
Question:
Which of the following is a method for initiating the uptake of a plasmid into a target eukaryotic
cell? A. Electroporation. B. Pillus. C. Lambdaphage viral vector. D. Heat shock in the presence of
CaCl2. E. Both A and D.
Answer:
E. Both A and D. The correct answer is E. Bacteria utilize a pillus for conjugation, in order to
exchange genetic material. The lambdaphage viral vector can be used injecting a specific sequence
of DNA into a target cell, but not for encouraging the cell to uptake plasmid DNA. Electroporation
involves using a small charge to cause the cell membrane to become more porous and uptake the
plasmid. Heat shock involves heating up the target cells to ~42ºC after keeping them on ice in the
presence of CaCl2, the cell membranes will become more porous and uptake the plasmid as with
electroporation.
Question:
Which is the best leaving group in a substitution reaction of an alkyl halide from the following
choices? A. F- B. Cl- C. Br- D. I- E. All are equally good leaving groups.
Answer:
,D. I- As size increases, basicity decreases, meaning a species will be less likely to act as a base; that
is, the species will be less likely to share its electrons. Iodine is the largest in the series.
Question:
The hardest substance known is diamond. Which best describes the structure of diamond? A. A 3D
network lattice of covalently bonded carbon atoms. B. A 2D network lattice of covalently bonded
carbon atoms. C. A 3D network lattice of carbon ions held together by ionic bonds. D. A 3D
network lattice of carbon ions in a sea of mobile valence electrons held together by metallic bonds.
E. A 3D network lattice of carbon atoms held together by hydrogen bonding and dispersion forces.
Answer:
A. A 3D network lattice of covalently bonded carbon atoms. Diamond is a 3D network lattice of
covalently bonded carbon atoms, far right. Graphite, another allotrope of carbon, is a 2D network
lattice of covalently bonded carbon atoms, left. A 3D network lattice of carbon ions held together by
ionic bonds. Does not exist, as carbon does not form ions and elements cannot be ionic, only
compounds can. A 3D network lattice of carbon ions in a sea of mobile valence electrons held
together by metallic bonds. This description is more metallic and similar to the description of
graphite, which can conduct electricity, but not diamond. A 3D network lattice of carbon atoms held
together by hydrogen bonding and dispersion forces. Hydrogen bonding and dispersion forces are
intermolecular forces, although dispersion forces are everywhere, but very weak.
Question:
DNA polymerases catalyze the production of polynucleotide chains. Which of the following
statements holds true: A. DNA polymerases do not require a primer in order to initiate the reaction
B. DNA polymerases require a primer with a free 5'-end C. DNA polymerases require a primer
base-paired to the template with a free 5'-end D. DNA polymerases require a RNA primer
base-paired to the template and with a free 3'-end. E. DNA polymerases require a DNA primer
base-paired to the template and with a free 5'-end.
Answer:
D. DNA polymerases require a primer base-paired to the template and with a free 3′-end. DNA
polymerase can add dNTPs (deoxynucleoside triphosphates) only to a pre-existing oligonucleotide
chain that is annealed to the template DNA. The enzyme adds dNTPs on a free 3′-OH- group.
Primers in replication are RNA based. These primers are then cut out via endonuclease activity and
replaced with DNA nucleotides. Finally, DNA ligase fuses the gap between the fragments of DNA.
, Question:
A researcher is attempting to predict the number of lions in an African habitat based on previous
data. He has formed a regression equation which is: Y = 25 + 0.5X - 2ZX, where X is the number of
years since the first population sample, and Z is the death rate of the lions. If it has been 10 years,
and the death rate is 1 lion per year, what is the predicted number of lions in the habitat? A. 10 B. 12
C. 16 D. 18 E. 20.
Answer:
A. 10 We are given a regression equation, which is Y = 25 +0.5X - 2ZX, where X is 10 years and Z
is 1 lion per year. This means that the equation can be substituted as Y = 25 +0.5(10) - 2(10)(1) = 10
lions remaining after 10 years. Y=25+0.5X-2Z X=10 years, Z=1 lion death per year
Y=25+0.5(10)-2(10)(1)= 10 lions remaining
Question:
For a mixture of gases in a container, the total pressure is equal to __________ because of what
law? A. The pressure of the largest gas, First Law of Thermodynamics B. The pressure of the
smallest gas, First Law of thermodynamics C. The sum of the partial pressures of all gases, Dalton's
law of partial pressure D. The pressure of the greatest gas, Dalton's law E. The sum of the top two
gases in the mixture, Dalton's law of partial pressure.
Answer:
C. The sum of the partial pressures of all gases, Dalton's law of partial pressure In a mixture of
gases, the total gaseous pressure is the sum of the partial pressures of all gases, according to Dalton's
law of partial pressure.
Question:
The first step of glycolysis is the phosphorylation of one molecule of glucose to generate glucose
6-phosphate. This reaction: A. is reversible. B. occurs in anaerobic organisms only. C. prevents
glucose from leaking through the plasma membrane. D. takes place in the inner membrane of
mitochondria. E. takes place in the outer membrane of mitochondria.
Answer:
C. prevents glucose from leaking through the plasma membrane. Glycolysis takes place in the
cytosol. It is a characteristic of eukaryotic and prokaryotic cells. The first reaction of glycolysis is
irreversible since it requires an energy investment. Glucose-6- phosphate cannot pass back through
the plasma membrane.
Question:
Mark's car began a journey from a point at 7 AM, traveling at 30 mph. At 8 AM Jason's car started
traveling from the same point at 40 mph in the same direction as Jason. At what time will Jason pass
Mark? A. 11 AM B. 10 AM C. 10:30 AM D. 11:30 AM E. Jason never passes Mark.
Answer:
A. 11 AM Let x be the number of hours it takes Jason to pass Mark. During this time, Jason travels
40x miles. Mark started an hour earlier, so he has driven for x+1 hours, during which he travels
30(x+1) miles. Therefore, 40x = 30(x+1) x=3 3 hours from 8 AM is: 11 AM.
Question:
Which of the following is a method for initiating the uptake of a plasmid into a target eukaryotic
cell? A. Electroporation. B. Pillus. C. Lambdaphage viral vector. D. Heat shock in the presence of
CaCl2. E. Both A and D.
Answer:
E. Both A and D. The correct answer is E. Bacteria utilize a pillus for conjugation, in order to
exchange genetic material. The lambdaphage viral vector can be used injecting a specific sequence
of DNA into a target cell, but not for encouraging the cell to uptake plasmid DNA. Electroporation
involves using a small charge to cause the cell membrane to become more porous and uptake the
plasmid. Heat shock involves heating up the target cells to ~42ºC after keeping them on ice in the
presence of CaCl2, the cell membranes will become more porous and uptake the plasmid as with
electroporation.
Question:
Which is the best leaving group in a substitution reaction of an alkyl halide from the following
choices? A. F- B. Cl- C. Br- D. I- E. All are equally good leaving groups.
Answer:
,D. I- As size increases, basicity decreases, meaning a species will be less likely to act as a base; that
is, the species will be less likely to share its electrons. Iodine is the largest in the series.
Question:
The hardest substance known is diamond. Which best describes the structure of diamond? A. A 3D
network lattice of covalently bonded carbon atoms. B. A 2D network lattice of covalently bonded
carbon atoms. C. A 3D network lattice of carbon ions held together by ionic bonds. D. A 3D
network lattice of carbon ions in a sea of mobile valence electrons held together by metallic bonds.
E. A 3D network lattice of carbon atoms held together by hydrogen bonding and dispersion forces.
Answer:
A. A 3D network lattice of covalently bonded carbon atoms. Diamond is a 3D network lattice of
covalently bonded carbon atoms, far right. Graphite, another allotrope of carbon, is a 2D network
lattice of covalently bonded carbon atoms, left. A 3D network lattice of carbon ions held together by
ionic bonds. Does not exist, as carbon does not form ions and elements cannot be ionic, only
compounds can. A 3D network lattice of carbon ions in a sea of mobile valence electrons held
together by metallic bonds. This description is more metallic and similar to the description of
graphite, which can conduct electricity, but not diamond. A 3D network lattice of carbon atoms held
together by hydrogen bonding and dispersion forces. Hydrogen bonding and dispersion forces are
intermolecular forces, although dispersion forces are everywhere, but very weak.
Question:
DNA polymerases catalyze the production of polynucleotide chains. Which of the following
statements holds true: A. DNA polymerases do not require a primer in order to initiate the reaction
B. DNA polymerases require a primer with a free 5'-end C. DNA polymerases require a primer
base-paired to the template with a free 5'-end D. DNA polymerases require a RNA primer
base-paired to the template and with a free 3'-end. E. DNA polymerases require a DNA primer
base-paired to the template and with a free 5'-end.
Answer:
D. DNA polymerases require a primer base-paired to the template and with a free 3′-end. DNA
polymerase can add dNTPs (deoxynucleoside triphosphates) only to a pre-existing oligonucleotide
chain that is annealed to the template DNA. The enzyme adds dNTPs on a free 3′-OH- group.
Primers in replication are RNA based. These primers are then cut out via endonuclease activity and
replaced with DNA nucleotides. Finally, DNA ligase fuses the gap between the fragments of DNA.
, Question:
A researcher is attempting to predict the number of lions in an African habitat based on previous
data. He has formed a regression equation which is: Y = 25 + 0.5X - 2ZX, where X is the number of
years since the first population sample, and Z is the death rate of the lions. If it has been 10 years,
and the death rate is 1 lion per year, what is the predicted number of lions in the habitat? A. 10 B. 12
C. 16 D. 18 E. 20.
Answer:
A. 10 We are given a regression equation, which is Y = 25 +0.5X - 2ZX, where X is 10 years and Z
is 1 lion per year. This means that the equation can be substituted as Y = 25 +0.5(10) - 2(10)(1) = 10
lions remaining after 10 years. Y=25+0.5X-2Z X=10 years, Z=1 lion death per year
Y=25+0.5(10)-2(10)(1)= 10 lions remaining
Question:
For a mixture of gases in a container, the total pressure is equal to __________ because of what
law? A. The pressure of the largest gas, First Law of Thermodynamics B. The pressure of the
smallest gas, First Law of thermodynamics C. The sum of the partial pressures of all gases, Dalton's
law of partial pressure D. The pressure of the greatest gas, Dalton's law E. The sum of the top two
gases in the mixture, Dalton's law of partial pressure.
Answer:
C. The sum of the partial pressures of all gases, Dalton's law of partial pressure In a mixture of
gases, the total gaseous pressure is the sum of the partial pressures of all gases, according to Dalton's
law of partial pressure.
Question:
The first step of glycolysis is the phosphorylation of one molecule of glucose to generate glucose
6-phosphate. This reaction: A. is reversible. B. occurs in anaerobic organisms only. C. prevents
glucose from leaking through the plasma membrane. D. takes place in the inner membrane of
mitochondria. E. takes place in the outer membrane of mitochondria.
Answer:
C. prevents glucose from leaking through the plasma membrane. Glycolysis takes place in the
cytosol. It is a characteristic of eukaryotic and prokaryotic cells. The first reaction of glycolysis is
irreversible since it requires an energy investment. Glucose-6- phosphate cannot pass back through
the plasma membrane.
