1. Eukaryotes resemble prokaryotes in that both: (more than one answer possible;
ID: 53944 • Letter: 1
Question
1. Eukaryotes resemble prokaryotes in that both:
(more than one answer possible; mark all correct answers for full credit; points will be deducted for incorrect answers)
are multicellular.
use DNA as their genetic material.
contain nuclei.
have organelles.
have cell membranes.
2. Eukaryotes differ from prokaryotes in that ______.
(more than one answer possible; mark all correct answers for full credit; points will be deducted for incorrect answers).
prokaryotes are unicellular while eukaryotes are multicellular.
as a general rule and with few exceptions, prokaryotes do not have membrane-bound organelles, while eukaryotes have many membrane-bound subcellular compartments..
eukaryotes have most of their genetic material contained within membrane-bound nuclei while prokaryotes’ genetic material is not separated from the cytoplasm by a membrane.
prokaryotic cells are surrounded by peptidoglycan cell walls while eukaryotic cells are surrounded by phospholipid membranes.
eukaryotes use DNA as their genetic material, but prokaryotes use RNA.
Refer to the question above: Based on the available information, the most likely type of interaction that affects the relative strength of the affinity between A and X is:
van der Waals interactions
Covalent bonds
Hydrogen bonds
Electrostatic interactions
3 points
QUESTION 7
Arrange the following according to their relative energies at room temperature (from least energy to most energy)
- 1. 2. 3. 4.
van der Waals interactions
- 1. 2. 3. 4.
Thermal energy
- 1. 2. 3. 4.
The phospho-anhydride bond in ATP
- 1. 2. 3. 4.
Hydrogen bond
3 points
QUESTION 8
The constituents of biological membranes that form a barrier to the movement of hydrophilic compounds across the membrane are ____________.
peripheral membrane proteins.
integral membrane proteins.
carbohydrates.
nucleic acids.
phospholipids.
3 points
QUESTION 9
The constituents of biological membranes that are most important in facilitation of movement of hydrophilic compounds across the membrane are ____________.
integral membrane proteins.
peripheral membrane proteins.
carbohydrates.
phospholipids.
nucleic acids.
3 points
QUESTION 10
Which of the following is NOT involved in lipid bilayer formation?
van der Waals interactions
a decrease in G.
electrostatic interactions
a decrease in entropy
the hydrophobic effect
3 points
QUESTION 11
Among the common features of the biosynthetic pathways of the three fundamental polymers of life, DNA, RNA, and protein, one can count their directionality. This means that:
All of these processes are guided or “directed” by another linear polymer.
All of these processes and all the organisms evolved from a single progenitor in a “direction” from simple to complex.
During synthesis the monomers are added sequentially “head to tail” rendering the molecule with a “direction”.
All of these processes can follow only one “direction” – from monomers to polymers.
3 points
QUESTION 12
A phospho-ester bond links together:
The ribose 1’ carbon and the nitrogenous base in RNA.
The 5’ phosphate group and 3’ hydroxyl group of sequential nucleotides of a polymer of RNA.
The phosphate group and the adenine moiety of ATP.
The 5’ phosphate group and 3’ hydroxyl group of the two antiparallel strands of a duplex DNA
3 points
QUESTION 13
A segment of DNA that contains no centromere will fail to be propagated efficiently in eukaryotic cells after mitosis because:
The DNA will be randomly segregated into daughter cells.
The DNA will not bind to the chromosome scaffold proteins.
The DNA will not bind helicase.
The replicated DNA will contain excess mutations.
The ends of the DNA will not be fully replicated and will be shortened.
3 points
QUESTION 14
Crossing of a diploid wild-type that is homozygous for gene A with a mutant that is heterozygous for a recessive mutation in gene A will result in F1 progeny, of which:
Three fourths show the mutant phenotype and one-fourth show the wild-type phenotype.
One-fourth show the mutant phenotype and three-fourths show the wild-type phenotype.
Half show the mutant phenotype and half show the wild-type phenotype.
All show the mutant phenotype.
All show the wild-type phenotype.
3 points
QUESTION 15
Cytidine triphosphate is: (more than one answer possible; mark all correct answers for full credit; points will be deducted for incorrect answers).
a component of DNA
a nucleoside
a precrusor of RNA
a nucleotide
3 points
QUESTION 16
DNA synthesis is restricted to the S-phase of the cell cycle. At which step during DNA replication is the process regulated?
DNA primase binding to allow initiation of primer synthesis.
Binding of proliferating cell nuclear antigen to DNA.
Elongation of the primer by DNA polymerase .
Binding of helicase complexes at origins of replication.
3 points
QUESTION 17
Dideoxy chain-termination sequencing reactions contain:
Only ddNTPs but no dNTPs.
Dideoxyribonucleoside triphosphates (ddNTPs) in equimolar concentration with deoxyribonucleoside triphosphates (dNTPs).
Both ddNTPs and dNTPs, with ddNTPs in high excess.
Both ddNTPs and dNTPs, with dNTPs in high excess.
3 points
QUESTION 18
During DNA replication, “lagging strand” synthesis: (more than one answer possible; mark all correct answers for full credit; points will be deducted for incorrect answers).
involves multiple protein factors.
is discontinuous, requiring synthesis of short “Okazaki fragments”.
All are correct
proceeds toward the replication fork.
occurs in a 3’ --> 5’ direction.
3 points
QUESTION 19
H-bonding is an important feature of the relatively flexible “B” form duplex DNA helix and the more rigid -helix found in many proteins. What causes the difference in flexibility of these helices?
The direction of the H-bonds in the protein -helix produces force that is parallel to the long axis of the helix; while the force is perpendicular to the long axis in B-DNA.
Proteins that bind to B-DNA increase its flexibility.
The H-bonds in the protein -helix have a higher energy than those in B-DNA.
Greater number of H-bonds in the protein -helix produces more rigid structure.
3 points
QUESTION 20
Important features of B form DNA include all EXCEPT:
G and C bases on complementary strands share 3 covalent bonds.
The two double helix strands are oriented antiparallel .
There are ~10.1 base pairs per turn of the helix.
The duplex can bend along the long axis.
Nitrogenous bases project toward the interior of the double helix.
3 points
QUESTION 21
DNA synthesis is restricted to the S-phase of the cell cycle. At which step during DNA replication is the process regulated?
Elongation of the primer by DNA polymerase .
DNA primase binding to allow initiation of primer synthesis.
Binding of helicase complexes at origins of replication.
Binding of proliferating cell nuclear antigen to DNA.
3 points
QUESTION 22
Protein factors that participate in DNA replication include:
Helicase.
DNA polymerase alpha.
DNA polymerase delta.
B and C
All of the above.
3 points
QUESTION 23
Initiation of DNA synthesis in E. coli requires all EXCEPT:
DNA polymerase
T7 RNA polymerase
a primer
a template strand
3 points
QUESTION 24
Monitoring the change in absorption of 260 nm wavelength light while elevating the temperature of the solution is a procedure that can be used to:
Determine the concentration of RNA.
Amplify specific DNA sequences to very high copy number.
Determine the melting temperature (Tm) of a specific DNA sequence.
Evaluate the linking number of supercoiled DNA.
Recently, the causative agent of a severe form of encephalitis (inflammation of the brain) was identified. Biochemical analysis of the pathogen revealed the presence of a single species of nucleic acid with the following base composition: A=20%, C=20%, G=30%, U=30%. Which of the following statement is most likely to be true?
The pathogen’s genetic material is single-stranded DNA.
The pathogen’s genetic material is single-stranded RNA.
The pathogen’s genetic material is double-stranded DNA
The pathogen’s genetic material is double-stranded RNA
3 points
QUESTION 30
DNA primase is required for initiation of DNA synthesis in vivo because: All of the above.
DNA primase can initiate synthesis of a polymer without a pre-existing nucleotide 3’ end.
DNA synthesis on the leading strand is discontinuous.
DNA primase can insert nucleotides without the need of a template.
DNA primase is more accurate than DNA polymerase.
3 points
QUESTION 31
The polymerase chain reaction (PCR) technique can be used for: (more than one answer possible; mark all correct answers for full credit; points will be deducted for incorrect answers).
Synthesis of large amounts of a specific segment of genomic DNA.
Non-templated chemical synthesis of oligonucleotides
End-tailoring of a DNA segment to introduce restriction endonuclease sites.
Synthesis of labeled DNA for use as hybridization probe.
3 points
QUESTION 32
Polymerase chain reaction (PCR):
Involves simultaneous elongation of 2 oligonucleotide primers bound at opposing positions on the complementary strands of a DNA template.
Produces multiple copies of RNA from a DNA template.
Involves simultaneous elongation of 2 oligonucleotide primers that bind to one strand of a DNA template.
Involves multiple cycles of formamide denaturation of DNA and primer annealing.
3 points
QUESTION 33
The two strands of the DNA double helix are oriented “antiparallel”, which means:
are multicellular.
use DNA as their genetic material.
contain nuclei.
have organelles.
have cell membranes.
Explanation / Answer
1. Eukaryotes resemble prokaryotes in having DNA as genetic material, have organelles and have cell membranes.
2. Eukaryotes differ from prokaryotes in that prokaryotes are unicellular while eukaryotes are multicellular, eukaryotes have most of their genetic material contained within membrane-bound nuclei while prokaryotes’ genetic material is not separated from the cytoplasm by a membrane, prokaryotic cells are surrounded by peptidoglycan cell walls while eukaryotic cells are surrounded by phospholipid membranes (a,c,d).
7. Hydrogen bond, Vanderwaals bond, thermal energy and Phosphoanhydride bond.
8. The answer is phospholipids.
9. The answer is integral membrane proteins.
10. The answer is decrease in entropy.
11.The answer is (b) All of these processes and all the organisms evolved from a single progenitor in a “direction” from simple to complex.
12. The answer is The 5’ phosphate group and 3’ hydroxyl group of sequential nucleotides of a polymer of RNA.
13. The answer is The DNA will not bind to the chromosome scaffold proteins.
14. The answer is Half show the mutant phenotype and half show the wild-type phenotype.
15. The answers are a nucleotide and a precursor for synthesis of RNA.
16. The answer is Binding of helicase complexes at origins of replication.
17. The answer is Both ddNTPs and dNTPs, with dNTPs in high excess.
18. The answes are discontinuous, requiring synthesis of short “Okazaki fragments” and proceeds toward the replication fork.
19. The answer is Greater number of H-bonds in the protein -helix produces more rigid structure.
20. The answer is Nitrogenous bases project toward the interior of the double helix.
21. The answer is Binding of helicase complexes at origins of replication.
22. The answer is all of the above.
23. The answer is T7 RNA polymerase
24. The answer is Determine the melting temperature (Tm) of a specific DNA sequence.
The answer is The pathogen’s genetic material is single-stranded RNA.
30. The answer is DNA primase can initiate synthesis of a polymer without a pre-existing nucleotide 3’ end.
31. The answers are Synthesis of large amounts of a specific segment of genomic DNA, End-tailoring of a DNA segment to introduce restriction endonuclease sites and Synthesis of labeled DNA for use as hybridization probe.
32. The answer is nvolves simultaneous elongation of 2 oligonucleotide primers bound at opposing positions on the complementary strands of a DNA template.
33. One strand runs from 5'-3' direction and another in 3'-5' direction.
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