BIO FINAL

What is the central dogma of molecular biology? a. Information passes from DNA to RNA to protein. b. For every A in DNA there is a T and for every G a C. c. DNA is a double stranded molecule that forms a helical structure. d. Regulation of gene expression is always more complex in eukaryotes than prokaryotes. e. Viral genomes are smaller than prokaryotic genomes which are smaller than eukaryotic genomes.

a. Information passes from DNA to RNA to protein.

Once translation starts, ribosomes are careful to move along mRNA molecules in three nucleotide steps. What would happen if ribosomes were to disregard that reading frame? a. Cells would have to have smaller genomes. b. Most proteins would take longer to translate. c. Frameshifts would introduce premature stop codons. d. Cells would make proteins that were too heat sensitive. e. Eukaryotic proteins would look more like their prokaryotic counterparts.

c. Frameshifts would introduce premature stop codons.

Which of the following functional group is most likely to impart a hydrophobic patch to organic compounds in which it is found? a. Phosphate b. Carboxyl c. Methyl d. Sulfhydryl e. Alcohol

c. Methyl

It will not be long before physicians routinely use information from the complete genome sequence of their patients. That information should allow the medicines that are proscribed to be tailored to an individual's particular metabolism so as to have the maximum effect with the smallest possible dose and side effects. What is this kind of individualized medicine called? a. Genomics. b. Metabolomics. c. Proteomics. d. Pharmacogenomics. e. Transcriptomics.

d. Pharmacogenomics

The discovery in 1974 of a group of enzymes in prokaryotes that break DNA chains at very specific sequences of nucleotides proved to be invaluable for researchers interested in both cloning and sequencing DNA. What are enzymes such as BamHI which cleaves DNA at the sequence 5'-GGATCC-3' called? a. Spliceosomes. b. Polymerases. c. Kinases. d. Dehydrogenases. e. Restriction enzymes.

e. Restriction enzymes.

Living systems are particularly sensitive to changes in hydrogen ion concentrations and utilize buffers extensively to minimize changes of pH within themselves. How many free H+ ions are present in one liter of neutral, distilled water. a. 1 x 10-7 moles. b. 7 moles c. 7 d. 7,000,000 e. 7 times as many OH- ions.

a. 1 x 10-7 moles.

The human genome is approximately 3.2 billion nucleotides long. 30.4% of those nucleotides are Adenine (A). What fraction of those nucleotides are Cytosine (C)? a. 19.6%. b. 39.2%. c. 30.4%. d. 55.8%. e. 60.8%

a. 19.6%.

The sum total of the DNA content within each of an organism's cells is often referred to as the organism's "C-value." It is reasonable to assume that C-value increases in a way that is proportional to organismal complexity. Which of the following organisms is most likely to have the smallest C-value? a. A virus. b. A bacteria. c. A human. d. A frog. e. A single-celled eukaryote (such as yeast).

a. A virus.

What is the central dogma of molecular biology? a. Information passes from DNA to RNA to protein. b. For every A in DNA there is a T and for every G a C. c. DNA is a double stranded molecule that forms a helical structure. d. Regulation of gene expression is always more complex in eukaryotes than prokaryotes.

a. Information passes from DNA to RNA to protein.

As result of Griffith's work, Alfred Hershey and Martha Chase radio-labelled viruses with radioactive isotopes of phosphorous (32P) and of sulfur (35S) in 1952. After allowing the viruses to interact with living bacterial cells for a short while, where did they find the 32P and 35S? a. Inside and outside the cells, respectively. b. Both were outside the cells. c. Outside and inside the cells, respectively. d. Both were inside the cells. e. Only in the air above the liquid the cells were growing in.

a. Inside and outside the cells, respectively.

Sir Alec Jefferies was the first person to actually see that the length of eukaryotic mRNAs was actually much shorter than the genes that coded for them when he hybridized mRNAs to genomic DNA and looked at the products with an electron microscope. What are the coding portions that are spliced out of the primary transcripts of eukaryotic genes? a. Introns. b. Exons. c. Xaptonuons. d. Open reading frames. e. Substitution regions.

a. Introns.

The chicken ovalbumin gene is expressed exclusively in hen oviduct cells. Which of the following is likely to be true? a. It would be in heterochromatin in most tissues but in euchromatin in oviduct cells. b. Its promoter would have a CpG island in oviduct cells but not in most other tissues. c. The C's of its CpG dinucleotides would be methylated in oviduct cells but not in most other tissues. d. The histones associated with it would be methylated and acetylated in most tissues but not in oviduct cells.

a. It would be in heterochromatin in most tissues but in euchromatin in oviduct cells.

The chicken ovalbumin gene is expressed exclusively in hen oviduct cells. Which of the following is likely to be true? a. It would be in heterochromatin in most tissues but in euchromatin in oviduct cells. b. Its promoter would have a CpG island in oviduct cells but not in most other tissues. c. The C's of its CpG dinucleotides would be methylated in oviduct cells but not in most other tissues. d. The histones associated with it would be methylated and acetylated in oviduct cells but not in most other tissues. e. It would be part of a looped domain that is heterochromatic in oviduct cells but euchromatic in most other tissues.

a. It would be in heterochromatin in most tissues but in euchromatin in oviduct cells.

The viruses that Alfred Hershey and Martha Chase radiolabelled with radioactive isotopes of phosphorous (32P) and of sulfur (35S) were T2 bacteriophage. T2 viruses, like the T4 viruses that destroy their host cell's genomic DNA as the first stage of their infection cycle, have an elaborate, "lunar-lander" like exterior when viewed with an electron microscope. Many viruses have less elaborate outer coats but from what bio-polymer are they all constructed? a. Proteins. b. Nucleic acids. c. Lipids. d. Sugars. e. Cellulose.

a. Proteins.

Agarose gel electrophoresis of DNA involves the movement of DNA in an electric field through the pores of a rectangular slab of agarose (a firm, gelatin-like material). Since DNA has a pronounced negative charge, it naturally moves toward positive electrodes when an electric current is applied to the gel. How do large pieces of DNA move relative to small ones in such a system? a. Slower because they become more tangled with the gel matrix. b. Faster because they have a greater number of negative charges. c. Slower because their covalent bonds have greater cumulative strength. d. Faster because they have a higher ratio of negative to positive charges. e. Faster because they form tighter tangles that can move through the gel more quickly.

a. Slower because they become more tangled with the gel matrix.

A chemical whose abbreviated name is IPTG is sufficiently similar to lactose to bind to the lac repressor bu it cannot be degraded by the enzyme β-galactosidase. What effect would the presence of IPTG have on the expression of the lac operon? a. The lac operon would be expressed at high levels even when there was no lactose. b. The lac operon would never be expressed (even when lactose levels are high). c. cAMP binding protein would always cause the lac operon to be expressed (even when lactose levels are low). d. The lac operon would never be expressed when lactose levels were high but always expressed when lactose levels were low. e. The lac operon would never be expressed when lactose levels were low but always expressed when lactose levels were high.

a. The lac operon would be expressed at high levels even when there was no lactose.

A chemical whose abbreviated name is IPTG is sufficiently similar to lactose to bind to the lac repressor but it cannot be degraded by the enzyme β-galactosidase. What effect would the presence of IPTG have on the expression of the lac operon? a. The lac operon would be expressed even when there was no lactose. b. The lac operon would never be expressed (even when lactose levels are high). c. cAMP binding protein would always cause the lac operon to be expressed (even when lactose levels are low). d. The lac operon would never be expressed when lactose levels were high but always expressed when lactose levels were low. e. The lac operon would never be expressed when lactose levels were low but always expressed when lactose levels were high.

a. The lac operon would be expressed even when there was no lactose.

Griffith found that Streptococcus pnemoniae that formed smooth colonies were capable of killing mice but that bacteria from rough colonies could not. What would he have found if he had infected mice with just the DNA of the smooth bacteria? a. The mice would not become ill. b. The mice would always die. c. The mice would develop tumors. d. The mice would be ill but not contagious. e. The mice would grow ill but always recover.

a. The mice would not become ill.

The binding site for the protein product of the lacI gene is normally found only once in the entire E. coli genome - the operator of the lac operon. What would happen if molecular biologists genetically engineered an E. coli genome in such a way as to include the nucleotide sequence of that binding site 50 nucleotides upstream of the promoter of the trp operon? a. There would be no effect on the expression of the trp operon. b. The lac operon's expression would be associated with tryptophan levels in the cell. c. The trp operon's expression would also be associated with lactose levels in the cell. d. The trp operon's expression would also be associated with glucose levels in the cell. e. The trp operon could never be expressed in those cells and they would die if tryptophan was not also included in their food.

a. There would be no effect on the expression of the trp operon.

Four different but complementary approaches were employed by the molecular biologists that obtained the complete nucleotide sequence of the human genome. Those approaches are: linkage mapping, physical mapping, sequencing and comparative sequence analysis. How do linkage and physical mapping differ? a. They use analyses of families and pieces of clones, respectively. b. Linkage determines the order of genes on chromosomes, physical mapping determines what features of our phenotypes are determined by those genes. c. Linkage describes the relationships between genes, physical mapping describes the relationship between organism and environment. d. Linkage mapping results in the construction of a "contig," while physical mapping generates a "haploid chromosome". e. Linkage mapping requires somatic cell hybrids while physical mapping requires flourescent in situ hybridization.

a. They use analyses of families and pieces of clones, respectively.

When populations of prokaryotic organisms coordinate their activities they can form communities that begin to rival the complexity of some eukaryotic organisms. How do such populations of bacteria determine how many bacteria are nearby? a. Through quorum sensing. b. Through aerobic respiration. c. Through plaque. d. Through countercurrent flow. e. Through anaerobic respiration.

a. Through quorum sensing.

In 1997 molecular biologists succeeded in replacing the nucleus of a sheep's egg cell with the nucleus of one of her mammary tissue cells. The resulting diploid cell was then coaxed to divide as if were a fertilized egg and a true clone was made. Why was it necessary to use a nucleus from relatively undifferentiated mammary tissue? a. Undifferentiated cells have more euchromatin. b. Undifferentiated cells have more heterochromatin. c. Undifferentiated cells live longer outside the body than do any others. d. The chromosomes of undifferentiated cells are frozen in their metaphase state and are easier to isolate. e. The chromosomes of cells in undifferentiated cells are in a polytene state and made the embryo better able to metabolize milk sugars.

a. Undifferentiated cells have more euchromatin.

Even though there are only four different nucleotides in DNA molecules, the specific order in which they are found is used by cells to spell out the order in which 20 different amino acids are used in making proteins. How many of the 64 possible triplet codes are reserved for use as stop codons? a. 1 b. 3 c. 24 d. 26 e. 44

b. 3

During the 1960's, the genetic code was deciphered by two different research groups. They found that ribosomes translated strings of poly-adenine (e.g. 5'-AAAAAAA-3') into polypeptides containing only the amino acid lysine. What is the nucleotide sequence of the anticodon of a lysine tRNA? a. 5'-Lys-3'. b. 5'-UUU-3'. c. 5'-TTT-3'. d. 5'-AAA-3'. e. 5'-UAG-3'.

b. 5'-UUU-3'.

Humans who are lactose intolerant no longer make the equivalent enzyme to beta galactosidase in the lac operon. Which of the following mutations to the lac operon of a prokaryotic cell might make it unable to synthesize beta galactosidase? a. A deletion of the operon's introns. b. A deletion of the start codon. c. A deletion of the lac repressor gene. d. Duplication of the lac repressor gene. e. Duplication of the lac permease gene.

b. A deletion of the start codon.

In the sixth of the ten steps of glycolysis, cells utilize an enzyme called glyceraldehyde phosphate dehydrogenase to enhance the speed of a reaction that would have occurred slowly even in its absence due to its favorable change in free energy. Which of the following best describes the reaction catalyzed by glyceraldehyde phosphate dehydrogenase? a. A six-carbon sugar is broken into two 3-carbon sugars. b. A three-carbon sugar is oxidized by the formation of NADH. c. A phosphate group is transferred from ATP to a six carbon sugar. d. A phosphate group is transferred from a 3-carbon sugar to ADP to form ATP. e. A double bond is formed in a 3-carbon molecule by removal of a water molecule.

b. A three-carbon sugar is oxidized by the formation of NADH.

By convention, the first nucleotide incorporated into an mRNA by an RNA polymerase is said to be the "+1" position for that gene. In prokaryotes, two relatively well-conserved sets of nucleotide sequences 35 and 10 base pairs (bp) upstream of that position are required for RNA polymerase to bind and be correctly oriented. How large are those Pribnow and TATA boxes? a. 2 and 3 bp, respectively. b. Both are 6 bp long. c. 6 and 18 bp, respectively. d. Both are 18 bp long. e. 18 and 12 bp, respectively.

b. Both are 6 bp long.

By convention, the first nucleotide incorporated into an mRNA by an RNA polymerase is said to be the "+1" position for that gene. In prokaryotes, two sets of nucleotide sequences 35 and 10 base pairs (bp) upstream of that position are required for RNA polymerase to bind and be correctly oriented. How many nucleotides get scrutinized in those -35 and -10 regions? a. 2 and 3 bp, respectively. b. Both are 6 bp long. c. 6 and 18 bp, respectively. d. Both are 18 bp long. e. 18 and 12 bp, respectively.

b. Both are 6 bp long.

When DNA is isolated from eukaryotic cells it is usually found to be tightly associated with a large amount of highly positively charged proteins called histones. The fact that histones are virtually identical across a wide variety of eukaryotic organisms implies that their function must be critically important. What is the complex of histones and DNA in eukaryotic nuclei called? a. Chondrocyte. b. Chromatin. c. Chromatid. d. Centriole. e. Centromere.

b. Chromatin.

As a general rule, eukaryotes are more complex than prokaryotes. Which of the following differences between prokaryotes and eukaryotes requires eukaryotic RNA polymerases and promoters to be substantially more complicated than their prokaryotic counterparts? a. Eukaryotic chromosomes have multiple origins of replication. b. Eukaryotes have to accomplish tissue-specific regulation of gene expression. c. Bacterial chromosomes have to hold so much information that they are the largest known. d. The genetic code on bacterial chromosomes is different from that on eukaryotic chromosomes. e. Bacterial chromosomes have long regions of non-coding sequence interrupting the coding regions of genes.

b. Eukaryotes have to accomplish tissue-specific regulation of gene expression.

When DNA is isolated from eukaryotic cells it is usually found to be tightly associated with a large amount of highly positively charged proteins called histones. The fact that histones are virtually identical across a wide variety of eukaryotic organisms implies that their function must be critically important. Which of the following is a function of histones? a. DNA repair (through Watsonian processing). b. Gene regulation (through chromatin packaging). c. Purine metabolism (through pyrimidine recycling). d. Cell division (through chromosome condensation). e. Recombination (through synaptonemal complexes).

b. Gene regulation (through chromatin packaging).

Most biological pathways require cells to possess two or more different enzymes in equal quantities for the efficient conversion of a starting material into an end-product. What feature of prokaryotes helps assure coordinated expression of genes in such pathways? a. They all use the same compounds as substrates. b. Genes with related functions are usually found in a single operon. c. The citric acid cycle converts excess proteins into more useful ones. d. Repressor proteins prevent more than one transcript being made at a time. e. Ribosomes only translate proteins when repressor proteins bind to their mRNA.

b. Genes with related functions are usually found in a single operon.

The wheat genome is about five times larger than the human genome. It may contain more protein coding regions (genes) as well. How can it be that humans are able to make many more different proteins than wheat? a. Most wheat genes are derived from stranded retroviruses. b. Humans do a lot more alternative splicing than wheat does. c. Humans use more amino acids to make their proteins than wheat does. d. Humans steal a wide variety of proteins from other organisms, including wheat. e. Wheat has mitochondria and chloroplasts while humans have only mitochondria.

b. Humans do a lot more alternative splicing than wheat does.

The cloning of a sheep, Dolly, from one of her own mammary gland cells in 1997 made possible many new opportunities for molecular biology. Which of the following would not be true of a human clone? a. It would look like a younger twin of its "parent." b. It would have the same memories as its "parent." c. It would have the same DNA profile as its "parent." d. It would have the same genetic defect as its "parent." e. It would have the same predisposition to genetic diseases as its "parent."

b. It would have the same memories as its "parent."

Which of the following would not be true of a cloned human? a. It would look like a younger twin of its "parent." b. It would have the same memories as its "parent." c. It would have the same DNA profile as its "parent." d. It would have the same genetic defect as its "parent." e. It would have the same predisposition to genetic diseases as its "parent."

b. It would have the same memories as its "parent."

The protein product of E. coli's trp-R gene can bind to the promoter of the trp operon but only when tryptophan is present. In that circumstance, RNA polymerases cannot transcribe the genes that code for the proteins that synthesize tryptophan. What type of regulation of gene expression is the trpR protein involved in? a. Positive regulation. b. Negative regulation. c. Conjugative regulation. d. Equivocal regulation. e. Esoteric regulation.

b. Negative regulation

A protein coded for by a gene immediately upstream of the lactose operon was found to specifically bind to the promoter in front of the beta-galactosidase gene but only when no lactose was present. When this protein was bound, RNA polymerase could not bind and begin transcription of the lactose operon. What type of regulation of gene expression is that protein involved in? a. Positive regulation. b. Negative regulation. c. Conjugative regulation. d. Equivocal regulation. e. Esoteric regulation. e. Esoteric regulation.

b. Negative regulation.

Viruses are so simple that they are often considered to not even qualify as being living things. In what type of macromolecule do they all store their genetic information? a. Proteins. b. Nucleic acids. c. Lipids. d. Sugars. e. Cellulose.

b. Nucleic acids.

Keri Mullis won a Nobel Prize for his description of what is now a very popular methodology for making billions of copies of a specific portion of a genome through repeated cycles of heating and cooling of the genome with a DNA polyermase, primers that flank the region of interest, and just a few additional reagents. What is this amplification process called? a. Reverse transcription (RT). b. Polymerase chain reaction (PCR). c. Northern hybridization (NH). d. Transcription (txn). e. Translation (tln).

b. Polymerase chain reaction (PCR).

Rosalind Franklin's X-ray crystal images told Watson and Crick that DNA had three important periodicities; one of 0.34 nm, one of 2 nm and one of 3.4 nm. They quickly realized that the first must correspond to the spacing between nitrogenous bases and that the second was probably related to the width of base pairing strands of DNA. What did the 3.4 nm measurement correspond to? a. The distance between proteins bound to DNA. b. The distance between turns on the helix that DNA forms. c. The length of chromosomes when they are packed into nuclei. d. The distance between phosphate groups in phosphodiester bonds. e. The size of protein molecules that bound to DNA to form chromatin.

b. The distance between turns on the helix that DNA forms.

Of all nature's molecules, nucleic acids are unique in their ability to direct their own replication. What feature(s) of DNA causes it to have this unusual characteristic? a. Their simple chemical composition can be edited by splicing reactions. b. The information contained in one strand is the same as that in the other. c. Ribosomes cannot interact with it directly because of the nuclear membrane. d. DNA has a very strong negative charge due to each nucleotide's phosphate group. e. Proteins are capable of recognizing differences in the nucleotide sequences of DNA.

b. The information contained in one strand is the same as that in the other.

Griffith found that Streptococcus pneumoniae that formed smooth colonies were capable of killing mice but that bacteria from rough colonies could not. What did he find when he infected mice with just the DNA of the smooth bacteria mixed with living rough bacteria? a. The mice would not become ill. b. The mice would always die. c. The mice would develop tumors. d. The mice would be ill but not contagious. e. The mice would grow ill but always recover.

b. The mice would always die.

F-factors are one of many different kinds of plasmids found in prokaryotes. Most plasmids confer a special ability upon their host cells such as immunity to a specific antibiotic. What makes cells with F-plasmids different from those without them? a. They are larger. b. They have F-pilli. c. They are auxotrophs. d. They are heterotrophs. e. They are autotrophs.

b. They have F-pilli.

If one strand of DNA has its nitrogenous bases arranged in this order: 5' - TACGAA - 3', what would be the order of nucleotides in its complementary strand? a. 5' - AAGCTA - 3'. b. 5' - ATGCTT - 3'. c. 5' - TTCGTA - 3'. d. 5' - AAGCAU - 3'. e. 5' - CAATTG - 3'.

c. 5' - TTCGTA - 3'.

In their initial models Watson and Crick tried to have nucleotides on the two strands of DNA base pair with themselves (i.e. G-G, and T-T). Why did they eventually reject that possibility? a. The polar charges at the tips of A's would drive them apart. b. DNA would not be complex enough to contain genetic information. c. A T-T base pair would not be wide enough to fit in the double helix. d. A G-G base pair would not be wide enough to fit in the double helix. e. An A-A base pair would cause too many hydrogen bonds to be formed.

c. A T-T base pair would not be wide enough to fit in the double helix.

Individuals with sickle cell anemia have a single difference in the amino acid sequence of their β-globin protein relative to people who do not have the disease. The specific mutation, a substitution, also resulted in the failure of a restriction enzyme to generate a particular fragment of DNA in afflicted individuals. What was that mutation? a. The transposition of the transcriptional terminator and transcriptional start sites in afflicted individuals. b. The substitution of the normal β-globin gene for a non-functional α-globin gene in afflicted individuals. c. A change that results in the incorporation of a valine instead of a glutamic acid as the sixth amino acid in the β-globin protein. d. The insertion of a single nucleotide into the very first codon of the β-globin gene of afflicted individuals relative to un-afflicted individuals. e. Deletion of a single, short stretch of DNA in afflicted individuals that corresponds to the promoter region of the β-globin gene in healthy individuals.

c. A change that results in the incorporation of a valine instead of a glutamic acid as the sixth amino acid in the β-globin protein.

F-factors are one of many different kinds of plasmids (extra-chromosomal pieces of DNA) found in prokaryotes. What exterior structure that is fundamentally important to bacterial sex do most of the 25 genes of a typical F-factor code for? a. Clathrin-coated pits. b. Erogenous zones. c. An F-pillus. d. Oligosaccharides. e. Glycoproteins.

c. An F-pillus.

Silicon (14Si) has only one electron in each of the four orbitals of its second level. Which of the atoms listed below would you expect to normally make the same number of covalent bonds as silicon? a. Hydrogen (1H) b. Sodium (11Na) c. Carbon (6C) d. Oxygen (8O) e. Magnesium (12Mg)

c. Carbon (6C)

Edward Southern was the first person to show that DNA molecules attached to paper-like membranes could specifically hybridize with other, free-floating DNA molecules that had been radio-labeled. Upon which of the following important features of DNA does Southern's technique rely? a. DNA replication is semi-conservative. b. RNA and DNA polymerases both add new nucleotides from 5' to 3'. c. Complementary strands of single stranded DNA have an affinity for each other. d. DNA has a very strong negative charge due to each nucleotide's phosphate group. e. DNA polymerase must first bind to double stranded DNA before it can begin DNA synthesis.

c. Complementary strands of single stranded DNA have an affinity for each other.

DNA is typically present in cells as a double helix consisting of two strands of polynucleotides. Why is the information on one strand of a DNA molecule redundant with that on its complementary strand? a. For every purine on one strand, there is a purine on the other. b. For every A and G on one strand, there is an A or G on the other. c. For every A and G on one strand, there is a T and C on the other. d. For every pyrimidine on one strand, there is a pyrimidine on the other. e. For every amino acid on one strand, there is a carbohydrate on the other.

c. For every A and G on one strand, there is a T and C on the other.

George Beadle and Edward Tatum used X-rays to alter the genetic material of a pink bread mold. Subsequent tests of the molds' ability to synthesize the materials it needed to grow revealed that the alteration of specific genes also altered specific enzymes. These observations were the basis of a theory that earned them a Nobel prize in 1958. What was that theory? a. Genes are made from aggregates of single proteins. b. Organisms must have more genes than polypeptides. c. For every one enzyme in an organism there was one gene. d. Mutations must always decrese the complexity of organisms. e. Mutations must always increase the complexity of organisms.

c. For every one enzyme in an organism there was one gene.

Which of the following lists things in increasing order of size? a. Operon, start codon, gene, nucleus. b. Oxygen, pyruvate, ethanol, glucose. c. Glucose, 5'UTR, gene, chromosome. d. Carbon dioxide, mRNA, tRNA, chloroplast. e. Nucleotide, ribose, chromosome, genome.

c. Glucose, 5'UTR, gene, chromosome.

Thomas Hunt Morgan in his studies with Drosophila was the first to realize that heritable traits (genes) were correlated with chromosomes. How was he able to reconcile the fact that Drosophila appeared to have a much larger number of traits than chromosomes? a. He found that each gene had its own chromosome during meiosis. b. He found that there were many more chromosomes than he first observed. c. He found that many traits were coded for on each of the four chromosomes. d. He found that each chromosome had a single gene that was highly pleiotropic. e. He found that the gene on each chromosome was expressed differently in every kind of cell.

c. He found that many traits were coded for on each of the four chromosomes.

In the first step of glycolysis a six-carbon sugar, glucose, interacts with ATP and, as a result, ADP and a six-carbon sugar with one phosphate group attached to it is formed. What is the name of the enzyme that catalyzes that reaction? a. Carbonic anhydrase. b. Phosphoglycerate isomerase. c. Hexokinase. d. Heme deoxygenase. e. Triosephosphate isomerase.

c. Hexokinase.

Many eukaryotic genomes contain vast tracts of non-coding or redundant DNA. Plants seem to have frequently accumulated numerous duplications of their whole genomes over the course of their evolutionary histories. What happens to a gene whose function is no longer needed because its copy serves its original purpose? a. It is quickly deleted to minimize the organism's genetic load. b. It undergoes recombination events with its copy that cause chromosomal rearrangements. c. Some acquire new functions that confer a selective advantage and come under functional constraint d. Its histones are heavily methylated and its cytosines are kept free of methylation in order to lessen its transcription levels. e. Gene conversion events between the copy and the original assure that any mutations to one will be corrected by the other.

c. Some acquire new functions that confer a selective advantage and come under functional constraint

DNA polymerases are notable enzymes both for their speed and their accuracy. Consider a nucleotide that is being added by a DNA polymerase to a growing polynucleotide chain. To which position on the ribose sugar of the nucleotide already present is a phosphodiester bond made? a. The 1' carbon. b. The 2' carbon. c. The 3' carbon. d. The 4' carbon. e. The 5' carbon.

c. The 3' carbon.

The ability to determine the order in which nucleotides appear within the DNA of our chromosomes was obviously a major advance in the study of biology since it allowed us to directly examine our genetic code. The first DNA sequencing method to be used won Maxam and Gilbert a Nobel Prize and was widely used until Sanger developed a new method roughly ten years later. How do Maxam and Gilbert's and Sanger's methods differ? a. The first is faster but not as accurate while Sanger's is accurate but slow. b. The first is slower but more accurate while Sanger's is faster but less accurate. c. The first involves chemical degradation of DNA while Sanger's is an enzymatic process. d. The first involves chemical synthesis of DNA while Sanger's is an enzymatic process. e. The first involves chemical synthesis of DNA while Sanger's is a translational process.

c. The first involves chemical degradation of DNA while Sanger's is an enzymatic process.

The binding site for the protein product of the lacI gene is normally found only once in the entire E. coli genome - the operator of the lac operon. What would happen if molecular biologists genetically engineered an E. coli genome in such a way as to include the nucleotide sequence of that binding site 50 nucleotides downstream of the promoter of the trp operon? a. There would be no effect on the expression of the trp operon. b. The lac operon's expression would be associated with tryptophan levels in the cell. c. The trp operon's expression would also be associated with lactose levels in the cell. d. The trp operon's expression would also be associated with glucose levels in the cell. e. The trp operon could never be expressed in those cells and they would die if tryptophan was not also included in their food.

c. The trp operon's expression would also be associated with lactose levels in the cell.

Arthur Kornberg won a Nobel prize for his work at Washington University in St. Louis that demonstrated that DNA polymerase is capable of making double stranded DNA out of a single stranded template. What other feature of DNA polymerases did he also discover in the course of his experimentation? a. That they were actually indistinguishable from the RNA polymerases responsible for transcription. b. That they were only capable of replicating DNA in the presence of both phospholipids and sugars. c. They must start with at least a short stretch of double stranded DNA as they add bases in a 5' to 3' direction. d. They were not capable of separating the two strands of DNA unless they were exposed to a pH of 8 or more. e. That other proteins interfered with their ability to replicate DNA unless sugars were also present.

c. They must start with at least a short stretch of double stranded DNA as they add bases in a 5' to 3' direction.

In both prokaryotes and eukaryotes the first amino acid incorporated into all new proteins is methionine. Two researchers found that prokaryotic ribosomes also required the sequence 5'-UAAGGAGG-3' (named the Shine-Delgarno sequence in their honor) 5' to a start codon in order to start translation. What causes ribosomes to release an mRNA and stop adding amino acids to a growing polypeptide? a. All prokaryotic proteins are 64 triplet codons long. b. When ribosomes encounter a polyadenylation signal they fall off mRNAs. c. When ribosomes encounter a codon for which there are no charged tRNAs they stop translating mRNAs. d. When a protein is large enough to assume its tertiary structure it forces ribosomes to fall off mRNAs. e. When cells no longer need a protein to be made, transcription factors bind to mRNA to make them unavailable to ribosomes.

c. When ribosomes encounter a codon for which there are no charged tRNAs they stop translating mRNAs.

Different Hfr strains exhibit different origins and directions of transfer of their bacterial chromosome. The order of transfer of genes close to the origin (O) of five different Hfr strains is given below. Hfr strain order of transfer of genes J O-lac-pro-ton-azi-leu K O-thi-met-ile-mtl-xyl-mal-str 9 O-ile-met-thi-thr-leu-azi AB219 O-gal-trp-his-str-mal-xyl-mtl AB221 O-pro-lac-ade-gal-trp-his Which of the following is a portion of the correct conjugational map of an E. coli chromosome? a. ile-mtl-met-thi-leu-thr. b. ile-met-thi-lac-pro-ton. c. thi-met-ile-mtl-xyl-mal. d. ile-met-mtl-thi-lac-thr. e. his-mtl-xyl-mal-str-trp.

c. thi-met-ile-mtl-xyl-mal.

pH and pOH are both related to each other and are both logarithmic scales. What would be the concentration of hydrogen ions in a solution whose pOH is 12? a. 10 moles/liter b. 1 mole/liter c. 0.1 mole/liter d. 0.01 mole/liter e. 1 x 10-12 moles/liter

d. 0.01 mole/liter

A hen is heterozygous for a sex-linked recessive lethal gene. What ratio of male to female chickens will be produced when she is crossed with a normal rooster? a. 3:1 b. 2:1 c. 1:1 d. 1:2 e. 1:3

d. 1:2

If one strand of DNA has its nitrogenous bases arranged in this order: 5' - CCATGA - 3', what would be the order of nucleotides in its complementary strand? a. 5' - GGTACT - 3'. b. 5' - GGUACU - 3'. c. 5' - AGTACC - 3'. d. 5' - TCATGG - 3'. e. 5' - CCATGA - 3'.

d. 5' - TCATGG - 3'.

Which of the following lists things in increasing order of size? a. Operon, start codon, gene, nucleus. b. Oxygen, pyruvate, ethanol, glucose. c. Glucose, ribose, gene, chromosome. d. Amino acids, tRNA, mRNA, ribosomes. e. Nucleotide, ribose, chromosome, genome.

d. Amino acids, tRNA, mRNA, ribosomes.

Which of the following is not an important difference between E. coli's single chromosome and the many chromosomes of human cells? a. E. coli's chromosome is circular and human chromosomes are linear. b. E. coli's chromosome is smaller than even the smallest human chromosome. c. The genetic code on E. coli's chromosome is the same as that on human chromosomes. d. Because E. coli is a simpler organism, it has a much lower density of information on its chromosome. e. Genes with related functions are clustered together on E. coli's chromosome but not on human chromosomes.

d. Because E. coli is a simpler organism, it has a much lower density of information on its chromosome.

Mitosis and the first phase of meiosis have many features in common. What is a fundamentally important difference between anaphase in mitosis and anaphase I of meiosis? a. Chromosomes are not as condensed in mitosis. b. Spindle fibers do not attach to kinetochores during meiosis. c. The nuclear envelope reforms only in the anaphase of mitosis. d. Chromatid pairs remain attached at their centromeres in meiosis. e. DNA is replicated one extra time at this stage of mitosis to make the cells diploid again.

d. Chromatid pairs remain attached at their centromeres in meiosis.

Matt Meselson and Frank Stahl grew cells with a heavy isotope of nitrogen (15N) so that their DNA would have greater density. After one generation of those cells were grown with a lighter isotope of nitrogen (14N) they found that all cells had DNA of intermediate density. Why did no cells have very dense or very light DNA? a. DNA replication is not conservative. b. DNA replication is ultra-conservative. c. DNA replication is conservative. d. DNA replication is semi-conservative. e. DNA replication is dispersive.

d. DNA replication is semi-conservative.

Which of the following is not an important difference between E. coli's single chromosome and the many chromosomes of human cells? a. E. coli's chromosome is circular and human chromosomes are linear. b. E. coli's chromosome is smaller than even the smallest human chromosome. c. The genetic code on E. coli's chromosome is the same as that on human chromosomes. d. E. coli's chromosome has long regions of non-coding sequence interrupting the coding regions of its genes. e. Genes with related functions are clustered together on E. coli's chromosome but not on human chromosomes.

d. E. coli's chromosome has long regions of non-coding sequence interrupting the coding regions of its genes.

As a general rule, eukaryotes are more complex than prokaryotes. Which of the following differences between prokaryotes and eukaryotes requires eukaryotic RNA polymerases and promoters to be substantially more complicated than their prokaryotic counterparts? a. Eukaryotic chromosomes have multiple origins of replication. b. Bacterial codons are two nucleotides long while eukaryotic codons are three. c. Bacterial chromosomes are linear while eukaryotic chromosomes are circular. d. Eukaryotes are multicellular organisms yet each cell has the same genetic instructions to work with. e. Bacterial chromosomes have long regions of non-coding sequence interrupting the coding regions of genes.

d. Eukaryotes are multicellular organisms yet each cell has the same genetic instructions to work with.

It will not be long before physicians routinely use information from the complete genome sequence of their patients. That information should allow the medicines that are proscribed to be tailored to the particular metabolism so as to have the maximum effect with the smallest possible dose and side effects. What is this kind of individualized medicine called? a. Genomics. b. Metabolomics. c. Proteomics. d. Pharmacogenomics. e. Transcriptomics.

d. Pharmacogenomics.

What one feature do all eukaryotic organisms have in common that no prokaryotic or archaebacterial organism has? a. Prokaryotes have no cilia but all eukaryotic cells do have them. b. All prokaryotes have chloroplasts whereas not all eukaryotes do. c. All prokaryotes have mitochondria whereas not all eukaryotes do. d. Prokaryotes have no internal membranes whereas all eukaryotes do. e. All prokaryotes have spindle pole bodies whereas not all eukaryotes do.

d. Prokaryotes have no internal membranes whereas all eukaryotes do.

Meselson and Stahl designed an experiment using two different isotopes of nitrogen (15N and 14N) that tested Watson and Crick's prediction that DNA replication occurred by a semi-conservative process. What would they have found after one round of cell divisions if DNA replication was actually conservative? a. No 15N would be found in any double stranded DNA molecules. b. No 14N would be found in any double stranded DNA molecules. c. All double stranded DNA molecules would contain equal mixtures of 15N and 14N. d. Some double stranded DNA molecules would contain only 15N and others would contain only 14N. e. A wide range of 15N and 14N compositions would be found in double stranded DNA molecules.

d. Some double stranded DNA molecules would contain only 15N and others would contain only 14N.

Unlike prokaryotes, eukaryotes extensively modify their mRNAs after they are first transcribed by RNA polymerases. Which of the following would not be included in the sequence of a processed eukaryotic mRNA? a. A translational start site. b. A translational stop site. c. A 5' cap. d. Spliced out intron sequences. e. The coding regions of a gene.

d. Spliced out intron sequences.

Once translation starts, ribosomes are careful to move along mRNA molecules in three nucleotide steps. What would happen if ribosomes were to disregard that reading frame? a. Cells would have to have smaller genomes. b. Most proteins would take longer to translate. c. Cells would make proteins that were too heat sensitive. d. The amino acids downstream of the frameshift would be incorrect. e. Eukaryotic proteins would look more like their prokaryotic counterparts.

d. The amino acids downstream of the frameshift would be incorrect.

Which of the following is a correct statement about the two DNA strands in a typical double stranded DNA molecule. a. One strand will contain genes; the other strand will contain promoter sequences. b. One strand will consist of only G's and A's; the other will contain only C's and T's. c. One strand will consist of only A's and T's; the other will contain only G's and C's. d. The two strands will be antiparallel to each other (the 5' end of one strand will match up with the 3' end of the other). e. One strand will always be made with a heavy isotope of nitrogren and the other will be made with a light isotope of nitrogen.

d. The two strands will be antiparallel to each other (the 5' end of one strand will match up with the 3' end of the other).

Which of the following is a correct statement about the two DNA strands in a typical double stranded DNA molecule? a. One strand will contain genes; the other strand will contain promoter sequences. b. One strand will consist of only A's and T's; the other will contain only G's and C's. c. One strand will consist of only G's and A's; the other will contain only C's and T's. d. The two strands will be antiparallel to each other (the 5' end of one strand will match up with the 3' end of the other). e. One strand will always be made with a heavy isotope of nitrogen and the other will be made with a light isotope of nitrogen.

d. The two strands will be antiparallel to each other (the 5' end of one strand will match up with the 3' end of the other).

A cross between two individuals who are AAbb and aaBB results in an F1 generation that displays the phenotype A_B_. The following phenotypes are obtained in an F2 cross: A_B_ = 110, A_bb = 16, aaB_ = 19, aabb = 15. What can be said about the A and B loci? a. bb must be lethal. b. aa must be lethal. c. They are genes Mendel reported. d. They are on the same chromosome. e. They are on different chromosomes.

d. They are on the same chromosome.

Complementary DNA sequences have a natural affinity for each other primarily because of their ability to form hydrogen between their nitrogenous bases. The three hydrogen bonds that form between guanines and cytosines and the two that form between adenines and thymines are known as "base pairs." How do these hydrogen bonds differ from the covalent bonds that connect a nitrogenous base to a sugar like deoxyribose in DNA? a. They take more energy to break because they involve bonds with hydrogen atoms instead of between nitrogen atoms. b. They take more energy to break because their energy is spread across two or three atoms instead of just one. c. They do not take as much energy to break because their energy is spread across two or three atoms instead of just one. d. They do not take as much energy to break because no sharing of electrons actually takes place

d. They do not take as much energy to break because no sharing of electrons actually takes place

Different Hfr strains exhibit different origins and directions of transfer of their bacterial chromosome. The order of transfer of genes close to the origin (O) of several different Hfr strains is given below. Hfr strain order of transfer of genes H O-azi-ton-pro-lac-ade 6 O-mtl-ile-met-thi-thr-leu-azi-ton-pro 4 O-thi-met-ile-mtl-xyl-mal-str AB311 O-his-trp-gal-ade-lac-pro-ton AB313 O-mal-str-his-trp-gal Which of the following is a portion of the correct conjugational map of a circular E. coli chromosome? a. ile-met-mtl-thi-leu-thr. b. xyl-mal-str-his-gal-trp. c. ile-mtl-met-thi-leu-thr. d. xyl-mtl-ile-met-thi-thr. e. ile-met-mtl-thi-thr-leu.

d. xyl-mtl-ile-met-thi-thr.

Over 200 enzymes such as BamHI and EcoRI have been found to cut DNA at specific nucleotide sequences and are commercially available to molecular biologists. Virtually all have been found to function as homo-dimers (two copies of the same protein working in concert) and to recognize palindromic DNA sequences. Which of the following is least likely to correspond to a site at which one of these enzymes cleaves DNA? ("N" means "any of the four nucleotides.") a. 5' - GGGCCC - 3'. b. 5' - AGATCT - 3'. c. 5' - GANTC - 3'. d. 5' - CCTNAGG - 3'. e. 5' - GCTACG - 3'.

e. 5' - GCTACG - 3'.

Analyses of lactose utilization by E. coli, carried out by François Jacob and Jacques Monod, provided the first major insights into bacterial gene regulation and resulted in the coining of the word "operon." Jacob and Monod found that the genes essential to lactose metabolism were in very close proximity and arranged on E. coli's chromosome in this order: beta-galactosidase, lactose permease, and transacetylase. What did these genes of the first studied operon have in common? a. A single attenuator sequence. b. A single translational start site. c. A single translational stop site. d. A single polyadenylation signal. e. A single promoter.

e. A single promoter.

Most of the energy extracted from glucose during glycolysis is initially harvested in the citric acid cycle as NADH. The high-energy electrons of NADH are used to create a proton gradient within mitochondria. Which of the following energy-rich compounds is the one that is synthesized as a direct result of that proton gradient? a. Glutamic acid. b. Glucose. c. Glyceraldehyde. d. Guanine. e. ATP.

e. ATP.

Sir Alec Jefferies was the first person to actually see that the length of eukaryotic mRNAs was actually much shorter than the genes that coded for them when he hybridized mRNAs to genomic DNA and looked at the products with an electron microscope. Introns can result in the transcription of millions of nucleotides into a eukaryotic RNA that are never translated. Which of the following is an advantage associated with introns? a. Viruses that insert into introns cannot become lytic. b. The transcriptional start-site for all eukaryotic genes are in introns. c. Splicing gives ribosomes a chance to catch up with fast moving RNA polymerases. d. Spliceosomes that remove introns also unwind double stranded DNA at replication forks. e. Alternative splicing allows for a single gene to give rise to different proteins in different tissues.

e. Alternative splicing allows for a single gene to give rise to different proteins in different tissues.

Most biological pathways require cells to possess two or more different enzymes in equal quantities for the efficient conversion of a starting material into an end-product. What feature of bacterial gene organization helps assure coordinated expression of genes in such pathways? a. They all use the same compounds as substrates. b. The citric acid cycle converts excess proteins into more useful ones. c. Repressor proteins prevent more than one transcript being made at a time. d. Ribosomes only translate proteins when repressor proteins bind to their mRNA. e. As part of single operons, they are all transcribed as a single polycistronic mRNA.

e. As part of single operons, they are all transcribed as a single polycistronic mRNA.

The burning of glucose releases 684 kcal per mole. The synthesis of glucose is just as energetically unfavorable as its combustion is favorable. How are living things able to make glucose synthesis a spontaneous reaction? a. By decreasing the amount of entropy outside of the cell. b. By capturing the heat energy that is released during the combustion of glucose. c. By allowing them to proceed in compartments in the cell with significantly different pHs and concentrations of sugars. d. By having most of the covalent bonds in glucose be made between atoms with similar electronegativities like hydrogen and carbon. e. By coupling the unfavorable energetics to chemical reactions that are energetically favorable like the conversion of ATP to ADP and Pi.

e. By coupling the unfavorable energetics to chemical reactions that are energetically favorable like the conversion of ATP to ADP and Pi.

RNases are enzymes that degrade mRNAs. Unlike ribosomes and RNA and DNA polymerases, RNases move 3' to 5' along nucleic acids. Long runs of A's at the 3' end of eukaryotic mRNAs serve as a protective barrier that makes it difficult for RNases to work through. When are those A's added to the 3' end of eukaryotic mRNAs? a. By ribosomes during translation. b. By the insertion of repetitive elements. c. By spliceosomes during mRNA splicing. d. By RNA polymerase during transcription. e. By polyadenylase during mRNA processing.

e. By polyadenylase during mRNA processing.

The different building blocks used to make each of the four kinds of biopolymers cause each kind of macromolecule to have a different function within living cells. What are the primary roles of carbohydrates, lipids, proteins and nucleic acids? a. Energy storage, catalysis, energy transfer and information storage, respectively. b. Information storage, energy storage, catalysis and energy transfer, respectively. c. Energy storage, energy transfer, information storage and catalysis, respectively. d. Catalysis, information transfer, energy storage and energy transfer, respectively. e. Energy transfer, energy storage, catalysis and information storage, respectively.

e. Energy transfer, energy storage, catalysis and information storage, respectively.

An average of three to four chiasma can be seen along the length of every human chromosome during prophase I of meiosis. Some regions appear to be less likely to form chiasma than others though. What effect would this have on recombination-frequency based gene mapping studies in these regions? a. Genes that seem to be dominant may really be recessive. b. Genes that seem to be recessive may really be dominant. c. Genes that seem to be dominant may really be only co-dominant. d. Genes that seem to be relatively far from each other may really be tightly linked. e. Genes that seem to be relatively close to each other may really be distantly linked.

e. Genes that seem to be relatively close to each other may really be distantly linked.

Edward Southern was the first person to show that DNA molecules attached to paper-like membranes could specifically hybridize with other, free-floating DNA molecules that had been radio-labeled. Upon which of the following important features of DNA does Southern's technique rely? a. DNA replication is semi-conservative. b. DNA polymerase can only extend double-stranded DNA molecules. c. RNA and DNA polymerases both add new nucleotides from 5' to 3'. d. DNA has a very strong negative charge due to each nucleotide's phosphate group. e. Hydrogen bonds can form between pairs of G's and C's and between pairs of A's and T's

e. Hydrogen bonds can form between pairs of G's and C's and between pairs of A's and T's.

Jacob and Monod won a Nobel prize in 1961 for their work on the lac operon. If a prokaryote had a mutant (non-functioning) version of the lacI gene but a normal copy of the lacZ gene, how would it respond to the presence and absence of lactose? a. It would never make beta galactosidase. b. It would only make beta galactosidase in the absence of lactose. c. It would only make beta galactosidase in the presence of lactose. d. It would only make beta galactosidase at temperatures above 37 C. e. It would make beta galactosidase both when lactose was present and absent.

e. It would make beta galactosidase both when lactose was present and absent.

Jacob and Monod won a Nobel prize in 1961 for their work with partial diploids and the lac operon. If a prokaryotic cell had a mutant (non-functioning) version of the lacI gene but a normal copy of the lacZ gene, how would it respond to the presence and absence of lactose? a. It would never make beta galactosidase. b. It would only make beta galactosidase in the absence of lactose. c. It would only make beta galactosidase in the presence of lactose. d. It would only make beta galactosidase at temperatures above 37 C. e. It would make beta galactosidase both when lactose was present and absent.

e. It would make beta galactosidase both when lactose was present and absent.

Each nucleotide is composed of two constant parts (a phosphate group and a pentose sugar) and one variable nitrogenous base. Which of the following is a not a difference between RNA and DNA? a. RNA uses uracil where DNA uses thymine. b. DNA is usually double stranded and RNA is usually single stranded. c. There is one fewer hydroxyl group on the ribose sugar of DNA than RNA. d. RNA is used to convey information to ribosomes and DNA is used to store information. e. Nucleotides are added to the 5' end of RNA and to the 3' end of DNA molecules.

e. Nucleotides are added to the 5' end of RNA and to the 3' end of DNA molecules.

Keri Mullis won a Nobel Prize for his description of what is now a very widely used methodology for making billions of copies of a specific portion of a genome through repeated cycles of heating and cooling of the genome with a DNA polymerase, primers that flank the region of interest, and just a few additional reagents. What is this amplification process called? a. Reverse transcription (RT). b. Translation (tln). c. Northern hybridization (NH). d. Transcription (txn). e. Polymerase chain reaction (PCR).

e. Polymerase chain reaction (PCR).

Diabetics generally need regular doses of the hormone insulin to survive. Which of the following represents a reasonable way of making a transgenic organism that could provide insulin for such treatments? a. Mixing just the DNA from the human insulin gene and smooth Streptococcus pnemoniae cells. b. Mixing just the DNA from the human insulin gene and rough Streptococcus pnemoniae cells. c. Mixing the DNA that codes for human insulin with cow's milk and feeding that mixture to immature cows. d. Putting the DNA sequence for the human insulin gene into a bacterial virus and using it to infect bacterial cells. e. Putting a bacterial promoter in front of the DNA sequence for the human insulin protein and inserting that construct as part of a plasmid into a bacterial cell.

e. Putting a bacterial promoter in front of the DNA sequence for the human insulin protein and inserting that construct as part of a plasmid into a bacterial cell.

By the 1940's, biologists knew that chromosomes carried genetic information and that they were composed of two substances: DNA and protein. Why did most researchers at that time feel that it was the proteins that were acting as the genetic material? a. Proteins alone could be used to transform cells. b. Many living things are made exclusively of protein. c. Agents that destroyed proteins in cells also prevented the cells from dividing. d. Because all living things had them but some living things did not have nucleic acids. e. They were found in all living things and their twenty amino acids seemed to be capable of coding for large amounts of information.

e. They were found in all living things and their twenty amino acids seemed to be capable of coding for large amounts of information.