MBB347 FINAL
All of the following statements are true EXCEPT _______. 1) Differences in ionic composition may exist between cells in different environments, but the external and internal ionic compositions of a cell within a certain environment are always identical to each other. 2) Differences in ionic composition across membranes can be used to drive molecular motor. 3) Differences in ionic composition across membranes can be generated through the action of ATP driven pumps. 4)Membrane potential associated with differences in ionic composition across membranes can be quickly and temporarily dissipated by transiently opening ion channels to allow rapid flow of certain ions according to their concentration gradient.
1) Differences in ionic composition may exist between cells in different environments, but the external and internal ionic compositions of a cell within a certain environment are always identical to each other.
Viruses are often not considered " living" because they have the following characteristics (True/False) : 1) They are acellular. 2) They are not made of organic molecules. 3) They lack membranes. 4) They lack nuclei. 5) They do not evolve. 6) They don't use DNA as their information storage molecule.
1) True 2) False 3) False 4) False 5) False 6) False ------------------------------ The only two true statements about viruses relating to them being considered "living" is the fact that they are acellular and rely on their host cells for providing their basic biological needs (all viruses rely on host translation for example). Viruses consist of organic molecules, of course. Some viruses are enveloped by a lipid bilayer (e.g. HIV-1, influenza). Being acellular, viruses obviously lack nuclei but prokaryotic cells also lack nuclei (and they are certainly "living"). Viruses certainly evolve and many rely on rather fast evolutionary rates for outsmarting their hosts (e.g. the evolution of drug resistance among HIV-1 isolates). Some viruses don't use DNA as their information storage molecule, but many others do
A 70-kg adult person could meet all of his/her entire energy needs for one day by consuming 3 moles of glucose (540 g - you are strongly advised, naturally, against such a diet!). Each molecule of glucose generates 30 ATP molecules when completely oxidized to CO2. The intracellular concentration of ATP is maintained constant in the cells at about 2 mM, and a 70-kg adult has about 25 liters of intracellular fluid. How many times per day, on average, does an ATP molecule turnover (= hydrolyzed and resynthesized)?
1,800 1 mole of glucose yields 30 moles of ATP 3 moles of glucose yield 90 moles of ATP. If there were no turnover (that is if ATP were not used by hydrolysis into AMP), then the concentration of ATP would rise to 90 moles/25 liters = 3.6 M = 3,600 mM. Since the intracellular ATP concentration remains constant at 2 mM that means that it continuously turning over: Turnover = 3,600 mM/ 2 mM = 1,800 times
Determine if the following features characterize eukaryotes, prokaryotes, both groups, or neither group. 1. They use a deoxyribonucleoside triphosphate as their "energy currency". 2. They chiefly use deoxyribonucleic acid molecules to store genetic information. 3. RNA molecules are their main catalysts. 4. Their DNA is typically found within sub-cellular organelles such as mitochondria, chloroplasts but mainly in the nucleus. 5. Their proteins consist of a common set of 20 different amino acids 6. Most of their genomes comprises protein-coding genes. 7. Their cell membranes chiefly comprise of phospholipid-based bilayer.
1. Neither prokaryotes nor eukaryotes 2. Both prokaryotes and eukaryotes 3. Neither prokaryotes nor eukaryotes 4. eukaryotes 5. Both prokaryotes and eukaryotes 6. prokaryotes 7. Both prokaryotes and eukaryotes ------------------------------------ A. Both EKs and PKs use ATP (or GTP for specialized functions) not dATP or dGTP. B. All cells use DNA for this purpose. C. While RNA catalyze some reactions, still most reactions are catalyzed by enzymes which are proteins.. D. DNA in EKs is found exclusively in the nucleus, in mitochondria and (in plants) plastids. These organelles are limited by double membranes. E. All cells do. F. The PK genome comprises mostly of protein-coding genes. This is in contrast to the EK genome, only a small portion of which encodes for proteins or functional RNAs. Most of EK DNA has no clear function and is dubbed (probably incorrectly) as "junk" DNA. G. All extant cells have bilayer whose main lipid components are phospholipids
What is the molecular weight of glucose?
180
A person whose blood glucose levels drop to below 70 mg/dL enters a serious hypoglycemic state. Hypoglycemia is potentially fatal if the glucose levels are not rapidly restored by consuming something loaded with sugar, like soda (for example sprite has 38 g sugar per a 12-oz can). How much Sprite should an 80-kg hypoglycemic man drink to raise his blood sugar level to normal? Express your answer in mL (rounded to the nearest mL). Please do not enter the units, though. You will need to make some simplifying assumptions. Assume that 45% of the sugar in sprite (infamous high fructose corn syrup) is glucose. Assume complete absorption, uniform distribution (in all aqueous compartments) and water content of 60% of body weight. Similarly ignore, elimination, metabolism etc. Remember: This is a very crude approximation and is not to be taken as a clinical recommendation.
202 1. First let's calculate by how much the blood glucose concentrations need to be raised to bring the levels to normal. [glucose]blood = 90 mg/mL -70 mg/mL = 20 mg/mL = 200 mg/L = 0.2 g/L 2. To translate this raise in concentration we need to know the volume. Our assumption treats the person as a barrel 60% of whose weight is water. Volume = 80 kg × 60% = 48 kg ⇒The volume of this amount of water is 48 L. 3. The mass of glucose we need is therefore Needed Glucose = 0.2 g/L × 48 L= 9.6 g 4. Next let's calculate what is the concentration of glucose in Sprite. [Glucose]Sprite = 0.45 × 38 g/can = 17.1 g/can = 17.1 g/12 oz = 1.425 g/oz [Glucose]Sprite = 1.425 g/oz = 1.425 g/30 mL = 0.0475 g/mL 5. Finally let's calculate what is volume of Sprite needed to give 9.6 g. Sprite = Needed Glucose / [Glucose]Sprite = 9.6 g / 0.0475 g/mL = 202 mL (~⅔ can)
DNA polymerases operate with high fidelity, but rarely may introduce a mutation by insertion of a mismatched nucleotide (does not base-pair correctly with the nucleotide base of the template). What feature of the enzyme allows proofreading to correct these errors? a) 3'->5' exonuclease activity. b) topoisomerase activity. c) helicase activity. d) 5'->3' polymerase activity.
3'->5' exonuclease activity.
In the US, the concentration of blood glucose is usually reported in milligrams (mg) per deciliter (dL = 100 mL = 0.1 L). A concentration of 90 mg/dL is considered the mean normal value. What is this value expressed in milimolar (mM) units (rounded to the nearest milimolar)?
5 Because 1 M = 1 mol/L, let us first calculate the normal blood sugar value in mg (or grams) per 1 liter: 90 mg/dL = 90 mg/0.1 L = 900 mg/L = 0.9 g/L Now, let us convert the mass (in g) to moles: If 180 g is 1 mol Then 0.9 g is x mole = 0.9 g • 1 mol/180 g = 0.005 mol Finally, therefore: [glucose] = 90 mg/dL = 0.005 mol/L = 0.005 M = 5 mM
You wish to use PCR to create a BamHI restriction endonuclease recognition site (GGATCC) at the 5' end of a target double stranded DNA sequence that begins 5' -ATG GGT CAT GAT TCG GAT CTC...-3', to allow it to be cut with BamHI and then ligated to another DNA fragment that has been cut likewise. Which primer will achieve this task? a) 5' -CCC GGA TCC ATG GGT CAT GAT TCG GAT CTC-3' b) 5' -ATG GGT CAT GAT TCG GAT GGA TCC-3' c) 5 - GGA TCC ATC CGA ATC ATG AGG CAT-3' d) 5-GAG ATC CGA ATC ATG AGG CAT GGA TCC-3'
5' -CCC GGA TCC ATG GGT CAT GAT TCG GAT CTC-3' The restriction enzyme recognition sequence must occur 5' of the primer sequence that matches the template. The 3' end of the primer must have a perfectly base-paired match with the template in order to efficiently prime DNA synthesis. Commonly, 2 or 3 extra nucleotides are added to the 5' end to increase the efficiency of endonuclease cleaveage of the PCR product.
You want to empirically determine the melting temperature of a linear double stranded DNA molecule, and decide to use the method of UV light absorption. You thus use a spectrophotometer to measure the absorption of 260 nm wavelength light by a solution of the DNA at different temperatures. The data are: (˚C, A260): 50, 0.20; 52, 0.20; 54, 0.22; 56, 0.24; 58, 0.26; 60, 0.28; 62, 0.28 What is the Tm for this DNA molecule? a) 60˚C b) There is not enough information to determine the Tm. c) 56˚C d) 52˚C
56˚C
Steps in polymerase chain reaction (PCR) include: a) All are correct. b) Annealing of 2 primers. c) Thermal denaturation of template DNA. d) Elongation of primers by a DNA polymerase.
All are correct.
Synthetic oligodeoxynucleotides can be used for: a) Hybridization probes for Southern blots. b) All are correct. c) Primers for use in DNA sequencing reactions. d) Primers for use in polymerase chain reaction (PCR).
All are correct.
Which structural property(s) of DNA are crucial for the conservation of genetic information? a) The ability to form a double helix. b) All are correct. c) Antiparallelism. d) Base-pair complementarity.
All are correct.
Restriction endonuclease recognition sequences are usually: a) All are correct. b) Greater than 10 bp in length. c) Sequences that are direct repeats. d) Also recognized by a DNA methylase.
Also recognized by a DNA methylase.
Living cells require input of free energy in order to be able to perform all of the following except _______. A) Incorrect to accumulate nutrients in the cell. B) Correct to breakdown macromolecules. C) to accumulate nutrients in the cell. D) to synthesize ATP from ADP and inorganic phosphate. E) transport ions across the membrane against their concentration gradient.
B) Correct to breakdown macromolecules.
DNA replication requires all EXCEPT: A) a DNA template. B) primers and deoxynucleotides. C) translation initiation factors. D) DNA polymerase.
C) translation initiation factors.
Which of the following sub-cellular structures contain DNA? A) lysosomes B) chloroplasts C) chromoplasts D) ribosomes E) mitochondria F) endoplasmic reticulum
Chloroplasts chromoplasts mitochondria
Review 23 on HW 1
Correct The polymerization reaction has ΔG < 0 at 37 ºC and ΔG > 0 at 0 ºC.
A suppressor mutation: a) Suppresses transcription of the affected gene. b) Activates the expression of reppressor proteins c) Is always a temperature-sensitive mutation. d) Counteracts the effects of a mutation in another gene.
Counteracts the effects of a mutation in another gene.
Which enzymes are used in the construction of a genomic DNA library? a) Taq DNA polymerase b) DNA ligase c) Reverse transcriptase d) Restriction endonuclease
DNA ligase Restriction endonuclease
Review Questions 5-9 (DNA % Problems)
HW 1
Which of the following enzymes will produce a 5' protruding single strand overhang? (The cut site is indicated by the * in the recognition sequence.): a) HindIII A*AGCTT b) SmaI CCC*GGG c) SacI GAGCT*C d) XhoI C*TCGAG
HindIII A*AGCTT XhoI C*TCGAG
In chain-termination sequencing of DNA, which is (are) important functions of the primer? Select all correct answers. a) It provides a nulceotide with a 5'-phosphate to which DNA polymerase can bond another nucleotide by its 3'-OH group. b) It anneals to the sequencing template DNA via base-pairing. c) It provides a nulceotide with a 3'-OH to which DNA polymerase can bond another nucleotide by its 5' phosphate. d) It serves as the substrate for DNA primase.
It anneals to the sequencing template DNA via base-pairing. It provides a nulceotide with a 3'-OH to which DNA polymerase can bond another nucleotide by its 5' phosphate.
Which of the following is true about a eukaryotic chromosome? a) It is a linear structure, although it may be folded in complex ways. b) It is composed of both DNA and proteins. c) It contains a single DNA molecule. d) It is replicated from a single origin of replication.
It is a linear structure, although it may be folded in complex ways. It is composed of both DNA and proteins. It contains a single DNA molecule.
Which of the following elements of a plasmid allows the host cell to synthesize new copies of the DNA? a) All are correct. b) Multiple cloning site. c) Selectable marker gene. d) Origin of replication.
Origin of replication.
An important functional feature of the structure of organic bases is that their geometry is: a) Polyhedral. b) Cuboidal. c) Tetrahedral. d) Planar
Planar
Which is a stage in the growth cycle of a lysogenic virus? a) Functional RT cannot be produced due to mutations in the viral gene that encodes RT. b) The host cell RT gene is not transcribed during viral infection. c) RT is not a normal constituent of the host cell cytoplasm. d) All are correct.
RT is not a normal constituent of the host cell cytoplasm.
A "phosphodiester bond" links together: a) Adenine and thymine in complementary strands of double stranded DNA molecule. b) The 3' hydroxyl group and the N9 atom of adenine in ATP. c) The α phosphate group and N9 atom of adenine in ATP. d) Sequential nucleosides in a molecule of DNA or RNA.
Sequential nucleosides in a molecule of DNA or RNA.
mRNA may be produced from a simple transcription unit or a complex one. The difference between simple and complex transcription units is: a) Simple transcription units occur in a homozygous condition for the allele, while complex units occur in a heterozygous condition. b) Simple transcription units are transcribed by RNA polymerase I, while complex units use RNA polymerase II. c) Complex transcription units undergo RNA splicing, while simple transcription units do not. d) Simple transcription units produce a single mRNA, while complex units can produce more than one mRNA.
Simple transcription units produce a single mRNA, while complex units can produce more than one mRNA.
mRNA may be produced from a simple transcription unit or a complex one. The difference between simple and complex transcription units is: a) Simple transcription units produce a single mRNA, while complex units can produce more than one mRNA. b) Simple transcription units occur in a homozygous condition for the allele, while complex units occur in a heterozygous condition. c) Simple transcription units are transcribed by RNA polymerase I, while complex units use RNA polymerase II. d) Complex transcription units undergo RNA splicing, while simple transcription units do not.
Simple transcription units produce a single mRNA, while complex units can produce more than one mRNA.
A linear DNA segment that contains a yeast-specific origin of replication and a centromere, but no telomere DNA is delivered into yeast cells. Which will you observe upon mitosis of transformed cells? a) The DNA is fails to replicate, but existing copies are randomly distributed to daughter cells. b) The DNA is replicated and distributed correctly to daughter cells. c) The DNA is replicated and randomly distributed to daughter cells. d)The DNA is replicated and distributed correctly to daughter cells, but is unstable and fails to be propagated over many generations.
The DNA is replicated and distributed correctly to daughter cells, but is unstable and fails to be propagated over many generations.
Which of the following is (are) factor(s) that determine melting temperature (Tm) of linear DNA in solution? a) The ion concentration of the solution. b) The base composition (frequency of occurrence of A, C, G, or T). c) The absorption of 260 nm UV light. d) The length of the DNA strands.
The Tm is a measure of the energy required to disrupt the H-bonding between base pairs that maintains double stranded structure of DNA. Absorption of UV light is used to distinguish linear and double stranded DNA, but does not have an effect on the hybridization behavior of DNA strands. a) The ion concentration of the solution. b) The base composition (frequency of occurrence of A, C, G, or T). d) The length of the DNA strands.
During mitosis, chromosomes are pulled to the two cell poles by the spindle. The spindle consists of filamentous "microtubules" which are made up of polymerized monomers of the protein tubulin. Tubulin has many hydrophobic amino acid side-chains on its surface (giving these surfaces overall hydrophobic nature) and the hydrophobic effect (so called "hydrophobic interactions") is the major force that holds the tubulin subunits together during polymerization and assembly of microtubules. Please explain what is the hydrophobic effect. A) The tendency of nonpolar molecules or parts of molecules to associate with each other in aqueous solution so as to minimize their direct interactions with water. B) A weak noncovalent interaction due to small, transient asymmetric electron distributions around atoms (dipoles). C) A noncovalent interaction between an atom (commonly oxygen or nitrogen) carrying a partial negative charge and a hydrogen atom carrying a partial positive charge. D) A weak noncovalent interaction due to small, transient asymmetric electron distributions around atoms (dipoles).
The tendency of nonpolar molecules or parts of molecules to associate with each other in aqueous solution so as to minimize their direct interactions with water. ------------------------- When hydrophobic molecules are dispersed in aqueous solutions, water molecules interfacing with the hydrophobic surface of the solute become ordered. This is an entropically unfavored state that is the driving force for the aggregation of the hydrophobic solute molecules thereby reducing surface area available for interaction of the water molecules, more water molecules are released from this highly ordered state and the entropy increases.
What is the maximal number of double covalent bonds that a carbon atom can make? Two One Three None Four
Two
Draw the structures of the bases that occur in DNA and/or RNA: adenine, guanine, cytosine, thymine, and uracil. Indicate the atoms that are H-bond donors and H-bond acceptors. Indicate the atom that is bonded to the 1' carbon of the ribose or deoxyribose ring. Now cut out the diagrams and arrange them to show correct base pairing between Watson-Crick base pairs in DNA, and the one alternative base pair in double stranded RNA. Yes, and I can prove it. Don't know. Maybe Yes, but my ferret shredded the evidence.
Yes, and I can prove it. ------------------------------- If you just chose the answer without actually doing the exercise, you should think about going back and doing it. The exercise will be very helpful in leading you to understand some of the structural and functional features of DNA, RNA, and DNA and RNA hybrids.
Consider the following single stranded DNA molecule with the sequence: 5'-GCTAGAAGTCTCGGAAAAACCGAGACTTCTAGC-3'. In aqueous solution of 50mM NaCl at 25˚C, the strand may form which structure(s)? a) A stem-loop with 14 base-pairs in the stem and a loop with 5 unpaired bases; or an antiparallel double-stranded helix with 28 base-pairs and a bubble in the center. b) A double-stranded helix with single stranded protrusions at both ends c) A stem-loop with 14 base-pairs in the stem and a loop with 5 unpaired bases. d) An antiparallel double-stranded helix with 28 base-pairs and a bubble in the center.
a) A stem-loop with 14 base-pairs in the stem and a loop with 5 unpaired bases; or an antiparallel double-stranded helix with 28 base-pairs and a bubble in the center.
You extract a mammalian cell culture sample to prepare total nucleic acids (DNA + RNA). Using this preparation, you produce complementary DNA (cDNA) by treating the nucleic acid preparation with reverse transcriptase, using oligo(dT) as the primer, in the presence of all four dNTPs. Using an aliquot of the cDNA as template, you then perform PCR using 2 primers that should amplify a DNA segment of 500 bp, based on the known sequence of the mRNA for this gene, which is transcribed from a simple transcription unit. When you electrophorese the PCR reaction on a 1% agarose gel, you do indeed observe a 500 bp product as expected. However, you also observe a 750 bp product. What could explain these results? a) Because genomic DNA was present in the total NA preparation, in addition to the expected 500 bp product, the PCR amplified a segment containing an intron lying between 2 exons. b) The primers were not specific for the target gene and also amplified a segment of cDNA from another gene. c) The oligo(dT) primer used to prime the reverse transcription reaction annealed with an oligo(A) sequence in the coding sequence a non-target mRNA. d) The target gene has a high proportion of G-C nucleotides in its sequence, which produces strong secondary structure in the mRNA.
a) Because genomic DNA was present in the total NA preparation, in addition to the expected 500 bp product, the PCR amplified a segment containing an intron lying between 2 exons. b) The primers were not specific for the target gene and also amplified a segment of cDNA from another gene.
In chain-termination sequencing of DNA, which of the following features enable or facilitate determination of the nucleotide sequence of the template? a) Elongation of an oliognucleotide primer by a DNA polymerase b) Determination of the size of the terminated chains. c) Termination of chain elongation by incorporation of dideoxynucleotides d) Determination of rate of elongation of DNA chains.
a) Elongation of an oliognucleotide primer by a DNA polymerase b) Determination of the size of the terminated chains. c) Termination of chain elongation by incorporation of dideoxynucleotides d) Determination of rate of elongation of DNA chains.
What enzyme catalyzes unwinding of the DNA helix during replication? a) Helicase. b) DNA polymerase alpha. c) Topoisomerase. d) DNA primase.
a) Helicase.
Guanosine monophosphate (GMP): a) Is a component of RNA. b) Contains a base that is found paired with adenine in double stranded DNA. c) Is a component of DNA and RNA. d) Is a pyrimidine nucleotide.
a) Is a component of RNA.
Some DNA polymerases contain a 3' -> 5' exonuclease activity, which enables: a) Removal of a mismatched nucleotide from the 3' end of a growing DNA chain b) Deamination of methylated cytosines. c) Excision of thymine dimers. d) Rapid and highly processive replication of the template DNA.
a) Removal of a mismatched nucleotide from the 3' end of a growing DNA chain
From "http://www.independent.co.uk/news/science/the-mouse-that-caught-a-cold-and-may-help-us-find-a-cure-777668.html" http://www.independent.co.uk/news/science/the-mouse-that-caught-a-cold-and-may-help-us-find-a-cure-777668.html , Monday, 4 February 2008: Scientists are claiming a significant breakthrough that could herald the development of new treatments and drugs to prevent the common cold. Being able to infect laboratory mice with rhinoviruses the main group of common cold viruses means scientists can now investigate how the virus infects an animal other than a human being. According to Professor Sebastian Johnston, a virologist at Imperial College London who led the mouse research, "We previously found that once inside the mouse cell a rhinovirus reproduces itself as well as it does in human cells. But the virus couldn't infect the mouse cell. The scientists modified the genes of the mice so that the cells lining their respiratory systems had a human version of: a) The cell surface receptor protein to which rhinovirus binds. b) Helicase, which is used to replicate the rhinovirus genome. c) The tRNA that carries lysine, an amino acid that is abundant in rhinovirus capsid protein. d) PCNA (proliferating cell nuclear antigen).
a) The cell surface receptor protein to which rhinovirus binds.
An enzyme that breaks DNA, dispels the tension, and reseals the strand ahead of a DNA replication growing fork is: a) Topoisomerase. b) Aminoacyl-tRNA synthetase. c) DNA polymerase. d) Phosphodiesterase.
a) Topoisomerase.
Considering the question above, one might be able to genetically modify the rhinovirus to enable it to infect a wild type mouse. Which experimental approach(es) might be feasible for this? Note that rhinovirus is a non-enveloped RNA virus that is related to poliovirus (Picoraviridae). a) Alter the viral membrane glycoprotein coding sequence so that the envelope can bind a surface molecule of the mouse respiratory epithelium cells. b) Alter the capsid protein coding sequence so that the capsid can bind to a surface molecule of the mouse respiratory epithelium cells. c) Insert a gene encoding a mouse epithelial cell membrane protein into the viral genome. d) Alter the viral RNA polymerase coding sequence so that the polymerase can replicate mRNA that encodes a mouse epithelial cell surface molecule.
b) Alter the capsid protein coding sequence so that the capsid can bind to a surface molecule of the mouse respiratory epithelium cells. ---------------------------------------- Theoretically, one could change the shape of the capsid protein so that it could bind to a surface molecule of the mouse respiratory epithelial cells, and thus effect entry. As the article states, once inside the host cell, the viral RNA replicates well. However, remember that changing the capsid protein shape may also alter its ability to assemble a capsid structure, so you would need to test your recombinant virus for this.
Measurement of the absorption of ultraviolet light (260nm) by a solution of DNA or RNA at a single temperature can be used to: a) Determine whether the DNA is linear or circular. b) Calculate the concentration of DNA or RNA in the solution. c) Determine whether the solution contains DNA, RNA, or both. d) Determine whether the DNA or RNA is single stranded or double stranded.
b) Calculate the concentration of DNA or RNA in the solution.
The DNA repair process that corrects mutations that occur during replication is called: a) nucleotide excision repair. b) mismatch excision repair. c) base excision repair. d) Non-homologous recombination.
b) mismatch excision repair.
During DNA replication, DNA synthesis on the "leading strand": a) proceeds away from the replication fork. b) proceeds toward the replication fork. c) requires a primer to initiate DNA synthesis. d) involves the ligation of Okazaki fragments.
b) proceeds toward the replication fork. c) requires a primer to initiate DNA synthesis.
What characteristic of DNA necessitates that some DNA synthesis is discontinuous, with formation and ligation of "Okazaki fragments"? a) DNA replication requires unwinding of the double helix. b)The 2 DNA strands are oriented antiparallel. c) all are correct d) DNA in the B form is a right-handed double helix.
b)The 2 DNA strands are oriented antiparallel.
The diameter of a typical bacteria cell can be all of the following except: 1) between 10-6 m and 2x10-6 m. 2) between 1,000 nM and 2,000 nM. 3) between 104 Å and 2x104 Å. 4) between 0.001 mm and 0.002 mm. 5) between 1.0 µm and 2.0 µm.
between 1,000 nM and 2,000 nM.
Which is a step in the replication cycle of a lytic virus? a) Transcription of virus-specific tRNAs. b) Integration of viral DNA into the host cell genome. c) Assembly of virus particles. All are correct.
c) Assembly of virus particles.
Which of the following is NOT required for both DNA replication and RNA transcription a) DNA. b) Proteins. c) Primers. d) RNA.
c) Primers.
To incorporate radiolabeled nucleotides into newly synthesized DNA, researchers use 32P-labeled dATP, in which the labeled P atom occupies the alpha, rather than the beta or gamma position, because: a) The radioactive alpha P atom in dATP lowers the activation energy for the synthetic reaction. b) The alpha P atom is more likely to undergo nuclear disintegration. c) The alpha P atom becomes incorporated into the DNA backbone. d) The beta and gamma P atoms become incorporated into the DNA backbone.
c) The alpha P atom becomes incorporated into the DNA backbone.
A DNA transposon that moves during DNA replication may produce what effect? a) Mitosis will result in aberrant segregation of chromosomes to daughter cells. b) After mitosis, both daughter cells will have an increased number of copies of that transposon. c) DNA replication will stop and mitosis will not occur. d) After mitosis, one daughter cell will have an increased number of copies of that transposon.
d) After mitosis, one daughter cell will have an increased number of copies of that transposon.
Which of the following are constituent(s) of deoxyribonucleotides? a) Organic bases b) Phosphate moieties c) Deoxyribose d) All are correct.
d) All are correct.
Why is DNA more stable than RNA? a) DNA replication is semiconservative. b) DNA can form a double-stranded helix. c) DNA lacks uracil. d) DNA lacks a 2'-OH group and therefore cannot undergo hydroxide ion-catalyzed hydrolysis.
d) DNA lacks a 2'-OH group and therefore cannot undergo hydroxide ion-catalyzed hydrolysis.
The ability of DNA to denature (complementary strands of the duplex separate) is important for which process? a) Cytokinesis during mitosis. b) Protein synthesis. c) DNA cleavage by endonucleases. d) DNA replication.
d) DNA replication.
Reverse transcriptase is an enzyme that: a) Transcribes ribosomal RNA genes. b) All are correct. c) Is an RNA polymerase. d) Makes a DNA copy of an RNA template.
d) Makes a DNA copy of an RNA template.
The pyrimidine bases found in DNA are: a) cytosine, thymine, and uracil b) adenine and guanine c) adenine and thymine d) Correct cytosine and thymine
d) cytosine and thymine
The polymerase chain reaction (PCR) technique can be used for: a) synthesis of RNA from genomic DNA. b) DNA integration into genomic DNA. c) All are correct. d) direct isolation of a specific segment of genomic DNA.
direct isolation of a specific segment of genomic DNA.
Crossing of a homozygous wild type with a mutant that is heterozygous for a dominant mutation will result in F1 progeny, of which: a) all show the wild-type phenotype. b) half show the wild-type phenotype and half show the mutant phenotype. c) three-fourths show the wild-type phenotype and one-fourth show the mutant phenotype. d) all show the mutant phenotype.
half show the wild-type phenotype and half show the mutant phenotype.
The constituents of biological membranes that are most important in facilitation of movement of hydrophilic compounds across the membrane are ____________. A) integral membrane proteins. B) nucleic acids. C) carbohydrates. D) phospholipids. E) peripheral membrane proteins.
integral membrane proteins.
Look at the structure of glucose in Figure 1. How should you calculate the molecular mass of glucose?
know how to find molecular mass look at structure of glucose 6x(atomic mass of C) + 6x(atomic mass of O) + 12x(atomic mass of H) (Question 17 on HW 1)
According to the endosymbiosis hypothesis, which of the following eukaryotic organelles evolved from originally independent prokaryotic ancestors? A) mitochondria B) chromoplasts C) ribosomes D) lysosomes E) Correct chloroplasts F) endoplasmic reticulum
mitochondria chloroplasts chromoplasts
What is the configuration around carbon atoms covalently binding four other partners? tetrahedral trigonal bipyramidal trigonal planar linear
tetrahedral
The synthesis of the three fundamental polymers of life, DNA, RNA and protein, is template-directed. This means that _____. 1) the system evolved in direction of more complexity from a 4-letter code of nucleotides to a 20-letter code of amino acids. 2) the monomers are assembled one by one in a sequence prescribed by an existing polymer. 3) all of them utilize a semi-conservative mode of synthesis. 4) DNA is copied from DNA, RNA is copied from RNA and protein is copied from protein.
the monomers are assembled one by one in a sequence prescribed by an existing polymer.
a) Genetic complementation can be used in yeast to determine: b) All are correct. c) whether two different recessive mutations are in the same or different genes. d) the mating type of a cell line. e) whether two different dominant mutations are in the same or different genes.
whether two different recessive mutations are in the same or different genes.
Assuming that ΔH and ΔS do not change with the temperature, are the values of ΔH and ΔS negative or positive? A) ΔH < 0 and ΔS > 0 B) ΔH > 0 and ΔS > 0 C) ΔH < 0 and ΔS < 0 D) ΔH > 0 and ΔS < 0
ΔH > 0 and ΔS > 0 ------------------------------------- A) If ΔH < 0 and ΔS < 0 ⇒ ΔH/ΔS > 0 At 37 ºC = 310 K ⇒ ΔG = ΔH - 310 K • ΔS < 0 ⇒ ΔH < 310•ΔS ⇒ ΔH/ΔS > 310 At 0 ºC = 273 K ⇒ ΔG = ΔH - 273 K • ΔS > 0 ⇒ ΔH > 273•Δ S ⇒ ΔH/ΔS < 273 But ΔH/ΔS can't be < 273 and > 310. Therefore, it can't be that both ΔH <0 and ΔS < 0. B) If ΔH < 0 and ΔS > 0 ⇒ ΔH/ΔS < 0 At 37 ºC = 310 K ⇒ ΔG = ΔH - 310 K • ΔS < 0 ⇒ ΔH < 310•Δ S ⇒ ΔH/ΔS < 310 At 0 ºC = 273 K ⇒ ΔG = ΔH - 273 K • ΔS > 0 ⇒ ΔH > 273• ΔS ⇒ ΔH/ΔS > 273 But ΔH/ΔS can't be < 0 and > 273. Therefore, it can't be that both ΔH < 0 and ΔS > 0. C) If ΔH > 0 and ΔS < 0 ⇒ ΔH/ΔS < 0 At 37 ºC = 310 K ⇒ ΔG = ΔH - 310 K • ΔS < 0 ⇒ ΔH < 310• ΔS ⇒ ΔH/ΔS > 310 At 0 ºC = 273 K ⇒ ΔG = ΔH - 273 K • ΔS > 0 ⇒ ΔH > 273• ΔS ⇒ ΔH/ΔS < 273 But ΔH/ΔS can't be < 273 and > 310. Therefore, it can't be that both ΔH > 0 and ΔS < 0. D) If ΔH > 0 and ΔS > 0 ⇒ ΔH/ΔS > 0 At 37 ºC = 310 K ⇒ ΔG = ΔH - 310 K • ΔS < 0 ⇒ ΔH < 310• ΔS ⇒ ΔH/ΔS < 310 At 0 ºC = 273 K ⇒ ΔG = ΔH - 273 K • ΔS > 0 ⇒ ΔH > 273• ΔS ⇒ ΔH/ΔS > 273 Therefore the values of ΔH and ΔS are such that 273 < ΔH/ΔS < 310, which is the only valid case. In other words, at the higher temperature the entropic component (TΔS) is large enough to make ΔG negative, but at lower temperatures, it is smaller than the enthalpic (ΔH) component making ΔG positive.
