Cell Biology Final Exam

Name three covalent modifications that can be made to an RNA molecule in eucaryotic cells before the RNA molecule becomes a mature mRNA.

1. A poly A tail is added. 2. A 5′ cap is added. 3. Introns can be spliced out.

A. If 0.5 mole of glucose weighs 90 g, what is the molecular weight of glucose? B. What is the concentration, in grams per liter (g/l), of a 0.25 M solution of glucose? C. How many molecules are there in 1 mole of glucose?

A. 180 daltons. A mole of a substance has a mass equivalent to its molecular weight expressed in grams. B. 45 g/l C. 6 × 1023 molecules

Which of the following statements about amino acids is true? (a) Twenty-two amino acids are commonly found in proteins. (b) Most of the amino acids used in protein biosynthesis have charged side chains. (c) Amino acids are often linked together to form branched polymers. (d) All amino acids contain an NH2 and a COOH group

Choice (d) is true. As their name implies, all amino acids have at least one amino (NH2) group and at least one acidic carboxylic (COOH) group. It is through these two groups that they form peptide bonds.

The genetic information of all cells is stored in their _____.

DNA

Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication. If false please explain. A) True B) False

False. Primase does not have a proofreading function, nor does it need one because the RNA primers are not a permanent part of the DNA. The primers are removed, and a DNA polymerase that does have a proofreading function fills in the remaining gaps

All functional DNA sequences inside a cell code for protein products. If the statement is false, explain why it is false. A) True B) False

False. Some sequences encode only RNA molecules, some bind to specific regulatory proteins, and others are sites where specific chrosomosomal protein structures are built (for example, centromeric and telomeric DNA)

The yeast GAL4 gene encodes a transcriptional regulator that can bind DNA upstream of genes required for the metabolism of the sugar galactose and turns them on. Gal4 has a DNA-binding domain and an activation domain. The DNA-binding domain allows it to bind to the appropriate sites in the promoters of the galactose metabolism genes. The activation domain attracts histone-modifying enzymes and also binds to a component of the RNA polymerase II enzyme complex, attracting it to the promoter so that the regulated genes can be turned on when Gal4 is also bound to the DNA. When Gal4 is expressed normally, the genes can be maximally activated. You decide to try to produce more of the galactose metabolism genes by overexpressing the Gal4 protein at levels fiftyfold greater than normal. You conduct experiments to show that you are overexpressing the Gal4 protein and that it is properly localized in the nucleus of the yeast cells. To your surprise, you find that too much Gal4 causes the galactose genes to be transcribed only at a low level. What is the most likely explanation for your findings?

For Gal4 to work properly, the DNA-bound Gal4 must attract histone-modifying enzymes and recruit RNA polymerase to the promoter. If there is too much Gal4 in the cell, the non-DNA-bound Gal4 (or free Gal4) will compete with the DNA-bound Gal4 for binding to histone-modifying enzymes and RNA polymerase. The excess amount of Gal4 forms nonproductive complexes with histone-modifying enzymes and RNA polymerase, preventing their recruitment to the promoter and lowering the level of transcription.

The following DNA sequence includes the beginning of a sequence coding for a protein. What would be the result of a mutation that changed the C marked by an asterisk to an A? 5′-AGGCTATGAATGGACACTGCGAGCCC....

The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the proteincoding sequence and in the correct reading frame (the beginning of the coding sequence is indicated by the ATG). Thus, translation would terminate after only four amino acids had been joined together, and the complete protein would not be made.

In principle, a eucaryotic cell can regulate gene expression at any step in the pathway from DNA to the active protein. Place the letter that corresponds with the correct control at the appropriate places on the diagram in the illustration. A) translation control B) transcriptional control C) RNA processing control D) protein activity control E) RNA transport and localization control

The figure is similar to Figure 8-3 in your book. Look there for the answers

The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this difference?

The gene contains one or more introns.

Why is the old dogma "one gene—one protein" not always true for eucaryotic genes?

The transcripts from some genes can be spliced in more than one way to give mRNAs containing different sequences, thus encoding different proteins. A single eucaryotic gene may therefore encode more than one protein.

There are five different nucleotides that become incorporated into a DNA strand. A) True B) False

There are five different nucleotides that become incorporated into a DNA strand. A) True B) False

From the sequencing of the human genome, we believe that there are approximately 24,000 proteincoding genes in the genome, for which there are an estimated 1500-3000 transcription factors (transcription regulatory factors). If every gene has a tissue-specific and signal-dependent transcription pattern, how can such a small number of transcriptional regulatory proteins generate a much larger set of transcriptional patterns?

Transcription regulator factors are generally used in combinations, thereby increasing the possible regulatory repertoire of gene expression with a limited number of proteins.

Although ____ contain the same type of molecules as cells, their inability to reproduce themselves by their own efforts means that they are not considered living matter.

Viruses

The relative strengths of covalent bonds and van der Waals interactions remain the same when tested in a vacuum or in water. However, this is not true of hydrogen bonds or ionic bonds, whose bond strength is lowered considerably in the presence of water in comparison with the bond strength observed in a vacuum. Explain these observations.

We estimate bond strengths by measuring the amount of energy needed to break them. In an aqueous solution, water may partly hydrogen bond with any charged or polar molecules, reducing the strength of the interaction they would otherwise have in the absence of water (in a vacuum). Covalent bonds and Van der Waals attractions have an intrinsic value that is independent of the environment.

procaryotic cells do not possess

a nucleus

Which protein(s) relate to discontinuous replication. A) primase B) single-strand binding protein C) sliding clamp D) RNA primers E) leading strand

a,b,c, and d

The same 20 ___ are used to make proteins.

amino acids

The core histones are small, basic proteins that have a globular domain at the Cterminus and a long extended conformation at the N-terminus. Which of the following is not true of the N terminal "tail" of these histones? A) It is subject to covalent modifications, B) It extends out of the nucleosome core. C) It binds to DNA in a sequence-specific manner. D) It helps DNA pack tightly. E) It contains lysine residues

c.)

The process of sorting human chromosomes pairs by size and morphology is called karyotyping. A modern method employed for karyotyping is called chromosome painting. How were individual chromosomes shown in a figure in the textbook "painted" as different colors? A) with a laser B) using fluorescent antibodies C) using fluorescent DNA molecules D) using green fluorescent protein E) by using Photoshop software

c.)

Which protein(s) relate to both continuous replication and discontinuous replication. A) primase B) single-strand binding protein C) sliding clamp D) RNA primers E) leading strand

c.)

Bright-field

employs a light microscope and requires that samples be fixed and stained in order to reveal cellular details.

Primase is needed to initiate DNA replication on both the leading strand and the lagging strand. If false please explain. A) True B) False

false

Transmission electron

has the ability to resolve cellular components a small as 2 nm

Use the terms listed to fill in the blanks in figure. A) A-T base pair B) G-C base pair C) deoxyribose D) phosphodiester bonds E) purine base F) pyrimidine base

images.

Drosophila melanogaster is a/an __________. This type of animal is the most abundant of all animal species, making it an appropriate choice as an experimental model.

insect

atomic number neutrons

is the number of protons/ electrons atomic mass -- atomic number

A. thaliana, or Arabidopsis, is a common weed. Biologists have selected it over hundreds of thousands of other flowering plant species to serve as an experimental model organism because __________________.

it can reproduce in 8-10 weeks it produces thousands of offspring per plant

Scanning electron

requires coating the sample with a thin layer of a heavy metal to produce threedimensional images of the surface of a sample

Chromosomes exist at different levels of condensation, depending on the stage of the cell cycle. If the statement is false, explain why it is false. A) True B) False

true

Eucaryotic chromosomes contain many different sites where DNA replication can be initiated. If the statement is false, explain why it is false. A) True B) False

true

Ionizing radiation and oxidative damage can cause DNA double-strand breaks. A) True B) False

true

Phase Contrast

uses a light microscope with an optical component to take advantage of the different refractive indexes of light passing through different regions of the cell.

Given the sequence of one strand of a DNA helix shown below, give the sequence of the complementary strand and label the 5′ and 3′ ends. 5′-GCATTCGTGGGTAG-3′,

5′-CTACCCACGAATGC-3′

Define a gene.

A gene is a segment of DNA that stores the information required to specify the particular sequence found in a protein (or, in some cases, the sequence of a structural or catalytic RNA).

A nucleosome contains two molecules each of histones ________ as well as of histones H3 and H4. A) H1 and H2A B) H2A and H2B C) H1 and H2B D) H1 E) H2A

A nucleosome contains two molecules each of histones H2A and H2B, as well as histones H3 and H4.

) The number of cells in an average-sized adult human is on the order of 1014. Use this information, and the estimate that the length of DNA contained in each cell is 2 m, to do the following calculations (look up the necessary distances and show your working): A. Over how many miles would the total DNA from the average human stretch?

A) 2 × 1014 m = 124,274,238,447 miles.

Answer the three questions (A, B, and C) after reading the following paragraph. Bacterial cells can take up the amino acid tryptophan from their surroundings, or, if the external supply is insufficient, they can synthesize trytophan by using enzymes in the cell. In some bacteria, the control of glutamine synthesis is similar to that of tryptophan synthesis, such that the glutamine repressor inhibits the transcription of the glutamine operon, which contains the genes that code for the enzymes required for glutamine synthesis. On binding to cellular glutamine, the glutamine repressor binds to a site in the promoter of the operon. A. Why is glutamine-dependent binding to the operon a useful property for the glutamine repressor? B. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells expressing a mutant form of the glutamine repressor that cannot bind to DNA? C. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells expressing a mutant form of the glutamine repressor that binds to DNA even when no glutamine is bound to it?

A) If sufficient glutamine is present in cells, the glutamine repressor will block the synthesis of enzymes that would make more glutamine. Similarly, if cells are starved for glutamine, the 5 unoccupied repressor would not bind to the DNA, and the enzymes that synthesize glutamine would be induced. These conditions permit a direct connection between the levels of glutamine and the expression of glutamine-synthesizing enzymes. B. The glutamine synthesis enzymes would be permanently switched on, regardless of the level of glutamine in the cells. C. The glutamine synthesis enzymes would be always switched off, regardless of the level of glutamine in the cells, because the repressor is always bound to the DNA. These cells will not be able to grow unless glutamine is added to the medium

When double-stranded DNA is heated, the two strands separate into single strands in a process called melting or denaturation. The temperature at which half of the duplex DNA molecules are intact and half have melted is defined as the Tm. A) Do you think Tm is a constant, or can it depend on other small molecules in the solution? B) Do you think high salt concentrations increase, decrease, or have no effect on Tm?

A) Tm depends on the identity and concentration of other molecules in the solution. B) High salt concentrations are more effective at shielding the two negatively charged phosphate-sugar backbones in the double helix from each other, so the two strands repel each other less strongly. Thus, a high salt concentration stabilizes the duplex and increases the melting temperature.

Which of the following is not a feature commonly observed in α helices? A) left-handedness B) one helical turn every 3.6 amino acids C) cylindrical shape D) amino acid side chains that point outward E) hydrogen bonds amongst the polypeptide backbone

A) left-handedness

) Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to carry? A) lysine B) glutamic acid D) leucine D) phenylalanine E) valine

A). As is conventional for nucleotide sequences, the anticodon is given reading from 5′ to 3′. The complementary base-pairing occurs between antiparallel nucleic acid sequences, and the codon recognized by this anticodon will therefore be 5′-AAG-3′.

You have two purified samples of protein Y: the wild-type (nonmutated) protein and a mutant version with a single amino acid substitution. When washed through the same gel-filtration column, mutant protein Y runs through the column more slowly than the normal protein. Which of the following changes in the mutant protein is most likely to explain this result? A) the loss of a binding site on the mutant protein surface through which protein Y normally forms dimers B) a change that results in the mutant protein's acquiring an overall positive instead of a negative charge C) a change that results in the mutant protein's being larger than the wild-type protein D) a change that results in the mutant protein's having a slightly different shape from the wild-type protein E) a change that results in the mutant protein's acquiring an overall negative instead of a positive charge

A). Dimers formed by a normal protein will run through the gel-filtration column faster than a mutant protein Y monomer. Choice B) and E) are unlikely, because gel-filtration columns separate proteins on the basis of size, not charge or affinity for small molecules. Choice C) is unlikely, because if the mutant protein were larger than normal it would be less able to enter the porous beads and would run through the column faster than the normal protein. Choice D) is unlikely, because a small change in shape without a change in size would be unlikely to have a major effect on the behavior of a protein in a gel-filtration column.

Below is the sequence from the 3′ end of an mRNA. 5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′ If you were told that this sequence contains the stop codon for the protein encoded by this mRNA, what is the anticodon on the tRNA in the P-site of the ribosome when release factor binds to the A-site? A) 5′-CCA-3′ B) 5′-CCG-3′ C) 5′-UGG-3′ D) 5′-UUA-3′ E) not enough information to determine the correct answer

A). The stop codon (UAA) is underlined in the mRNA sequence below; this is the only stop codon on this piece of mRNA. The codon (UGG) preceding the stop codon will be binding to a tRNA in the P-site of the ribosome when release factor binds to the A-site. The anticodon of the tRNA will bind to the codon UGG and will be CCA. 5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′

Indicate whether the statements below are true or false. If a statement is false, explain why it is false. A. Electrons are constantly moving around the nucleus of the atom, but they can move only in discrete regions. B. There is no limit to the number of electrons that can occupy the fourth electron shell. C. Atoms with unfilled outer electron shells are especially stable and are therefore less reactive. D. Covalent bonds are formed when electrons are either shared or transferred between atoms.

A. True. B. False. The fourth electron shell has the capacity to hold 18 electrons. C. False. Atoms that have their outer electron shells filled are the most stable and least reactive. Atoms with unfilled outer shells are more reactive because they tend to share or transfer electrons to fill and therefore stabilize the outer shell. D. False. Covalent bonds are formed exclusively when electrons are shared between atoms

A. Write out the sequence of amino acids in the following peptide, using the full names of the amino acids. Pro-Val-Thr-Gly-Lys-Cys-Glu B. Write the same sequence with the single-letter code for amino acids. C. According to the conventional way of writing the sequence of a peptide or a protein, which is the C-terminal amino acid and which is the N-terminal amino acid in the above peptide?

A. proline-valine-threonine-glycine-lysine-cysteine-glutamic acid (or glutamate) B. PVTGKCQ C. C-terminal is glutamic acid (or glutamate); N-terminal is proline.

List three ways in which the process of eucaryotic transcription differs from the process of bacterial transcription.

Any three of the following are acceptable. 1. Bacterial cells contain a single RNA polymerase, whereas eucaryotic cells have three. 2. Bacterial RNA polymerase can initiate transcription without the help of additional proteins, whereas eucaryotic RNA polymerases need general transcription factors. 3. In eucaryotic cells, transcription regulators can influence transcriptional initiation thousands of nucleotides away from the promoter, whereas bacterial regulatory sequences are very close to the promoter. 4. Eucaryotic transcription is affected by chromatin structure and nucleosomes, whereas bacteria lack nucleosomes.

For each of the following indicate whether the individual folded polypeptide chain forms a globular (G) or fibrous (F) protein molecule. A. keratin B. lysozyme C. elastin D. collagen E. hemoglobin F. actin (monomer)

A—F B—G C—F D—F E—G F—G

Which of the following statements about the genetic code is correct? A) All codons specify more than one amino acid. B) The genetic code is redundant. C) All amino acids are specified by more than one codon. D) All codons specify an amino acid. E) Codons encode for 21 different amino acids

B). Most amino acids can be specified by more than one codon. Each codon specifies only one amino acid (choice A)). Tryptophan and methionine are encoded by only one codon (choice C)). Some codons specify translational stop signals (choice D)).

The piece of RNA below includes the region that codes the binding site for the initiator tRNA needed in translation. 5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′ Which amino acid will be on the tRNA that is the first to bind to the A-site of the ribosome? A) methionine B) arginine C) cystine D) valine E) leucine

B). The initiator methionine is underlined on the RNA molecule below. 5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′ The first tRNA to bind at the A-site is the second codon of the protein, because the initiator tRNA is already bound to the P-site when translation begins. The codon that follows the binding site for the initiator tRNA is CGU, which codes for arginine.

Which of the following statements about the proteasome is false? A) Ubiquitin is a small protein that is covalently attached to proteins to mark them for delivery to the proteasome. B) Proteases reside in the central cylinder of a proteasome. C) Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold. D) The protein stoppers that surround the central cylinder of the proteasome use the energy from ATP hydrolysis to move proteins into the proteasome inner chamber. E) Proteases cleave the polypeptide backbone.

C) Once proteins are sent to the proteasome, proteases degrade them. Chaperone proteins provide a place for misfolded proteins to attempt to refold.

Fully folded proteins typically have polar side chains on their surfaces, where electrostatic attractions and hydrogen bonds can form between the polar group on the amino acid and the polar molecules in the solvent. In contrast, some proteins have a polar side chain in their hydrophobic interior. Which of following would not occur to help accommodate an internal, polar side chain? A) A hydrogen bond forms between two polar side chains. B) A hydrogen bond forms between a polar side chain and protein backbone. C) A hydrogen bond forms between a polar side chain and an aromatic side chain. D) Hydrogen bonds form between polar side chains and a buried water molecule. E) All (A-D)

C). Choices A), B), and D) are all common mechanisms employed to accommodate buried polar amino acids. Choice C) is not likely to accomplish this because aromatic side chains are nonpolar, hydrophobic residues and will not interact favorably with a polar, hydrophilic side chain.

In principle, how many different cell types can an organism having four different types of transcription regulator and thousands of genes create? A) up to 4 B) up to 8 C) up to 16 D) thousands E) millions

C). The type of cell is determined by the particular combination of transcription regulators active within it. With four different proteins available, there is one possibility with no proteins at all and one with all four proteins. There are four possibilities with one protein each, six possible combinations of two different proteins, and four possible combinations of three different proteins. 1. None 2. ABCD 3. A 4. B 5. C 6. D 7. AB 8. AC 9. AD 10. BC 11. BD 12. CD 13. ABC 14. ABD 15. BCD 16. CDA

DNA replication is considered semiconservative because _______________________. A) after many rounds of DNA replication, the original DNA double helix is still intact B) each daughter DNA molecule consists of two new strands copied from the parent DNA molecule C) each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand D) new DNA strands must be copied from a DNA template E) none of the above is (are) correct

Choice C) is the correct answer. Choices A) and B) are false. Although choice D) is a correct statement, it is not the reason that DNA replication is called semiconservative.

Several members of the same family were diagnosed with the same kind of cancer when they were unusually young. Which one of the following is the most likely explanation for this phenomenon? It is possible that the individuals with the cancer have _______________________. A) inherited a cancer-causing gene that suffered a mutation in an ancestor's somatic cells B) inherited a mutation in a gene required for DNA synthesis C) inherited a mutation in a gene required for mismatch repair D) inherited a mutation in a gene required for the synthesis of purine nucleotides E) inherited a mutation in a gene required for the synthesis of pyrimidine nucleotides

Choice C) is the correct answer. In fact, affected individuals in some families with a history of early-onset colon cancer have been found to carry mutations in mismatch repair genes. Mutations arising in somatic cells are not inherited, so choice A) is incorrect. A defect in DNA synthesis or nucleotide biosynthesis would probably be lethal, so choices B) and D) are incorrect.

Transcription is similar to DNA replication in that ___________________. A) an RNA transcript is synthesized discontinuously and the pieces are then joined together B) it uses the same enzyme as that used to synthesize RNA primers during DNA replication C) the newly synthesized RNA remains paired to the template DNA D) nucleotide polymerization occurs only in the 5′-to-3′ direction E) its products are always double stranded

Choice D) is correct. Choice A) is incorrect because an RNA transcript is made by a single polymerase molecule that proceeds from the start site to the termination site without falling off. The enzyme used to make primers during DNA synthesis is indeed an RNA polymerase, but it is a special enzyme, primase, and not the enzyme that is used for transcription, which is why choice B) is incorrect. Choice C) is false.

A molecule of bacterial DNA introduced into a yeast cell is imported into the nucleus but fails to replicate the yeast DNA. Where do you think the block to replication arises? Choose the protein or protein complex below that is most probably responsible for the failure to replicate bacterial DNA. Please explain your answer. A) primase B) helicase C) DNA polymerase D) initiator proteins E) A, B, and C

Choice D) is the correct answer. DNA from all organisms is chemically identical except for the sequence of nucleotides. The proteins listed in choices A), B) and C) can act on any DNA regardless of its sequence. In contrast, the initiator proteins recognize specific DNA sequences at the origins of replication. These sequences differ between bacteria and yeast

You wish to produce a human enzyme, protein A, by introducing its gene into bacteria. The genetically engineered bacteria make large amounts of protein A, but it is in the form of an insoluble aggregate with no enzymatic activity. Which of the following procedures might help you to obtain soluble, enzymatically active protein? Explain your reasoning. A) Make the bacteria synthesize protein A in smaller amounts. B) Dissolve the protein aggregate in urea, then dilute the solution and gradually remove the urea. C) Treat the insoluble aggregate with a protease. D) Make the bacteria overproduce chaperone proteins in addition to protein A. E) Heat the protein aggregate to denature all proteins, then cool the mixture.

Choices A, B, and D are all worth trying. Some proteins require molecular chaperones if they are to fold properly within the environment of the cell. In the absence of chaperones, a partly folded polypeptide chain has exposed amino acids that can form noncovalent bonds with other regions of the protein itself and with other proteins, thus causing nonspecific aggregation of proteins. A) Because the protein you are expressing in bacteria is being made in large quantities, it is possible that there are not enough chaperone molecules in the bacterium to fold the protein. Expressing the protein at lower levels might increase the amount of properly folded protein. B) Urea should solubilize the protein and completely unfold it. Removing the urea slowly and gradually often allows the protein to refold. Presumably, under less crowded conditions, the protein should be able to refold into its proper conformation. C) Treating the aggregate with a protease, which cleaves peptide bonds, will probably solubilize the protein by trimming it into pieces that do not interact as strongly with one another; however, chopping up the protein will also destroy its enzymatic activity. D) Overexpressing chaperone proteins might increase the amount of properly folded protein. E) Heating can lead to the partial denaturation and aggregation of proteins to form a solid gelatinous mass, as when cooking an egg white, and rarely helps solubilize proteins.

Which of the following statements about the newly synthesized strand of a human chromosome is true? A) It was synthesized from a single origin solely by continuous DNA synthesis. B) It was synthesized from a single origin by a mixture of continuous and discontinuous DNA synthesis. C) It was synthesized from multiple origins solely by discontinuous DNA synthesis. D) It was synthesized from multiple origins by a mixture of continuous and discontinuous DNA synthesis. E) It was synthesized by the use of ATP hydrolysis only.

D). Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins. Consider a single replication origin. The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis; the fork moving in the opposite direction synthesizes a portion of the same daughter strand discontinuously as part of lagging-strand synthesis.

You have a segment of DNA that contains the following sequence: 5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′ 3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′ If you know that the RNA transcribed from this segment contains the following sequence: 5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA-3′ Which of the following choices best describes how transcription occurs? A) The top strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′. B) The top strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′. C) The bottom strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′. D) The bottom strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′. E) Both the top and the bottom strands could be used as the template.

D). The bottom strand can hybridize with the RNA molecule and thus is the template strand. The polymerase moves along the DNA in a 3′ to 5′ direction, because the RNA nucleotides are joined in a 5′ to 3′ polarity

Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads to which the polynucleotide 5′-TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached. After a short incubation, the beads are then extracted from the mixture. When you analyze the cellular nucleic acids that have stuck to the beads, which of the following is most abundant? A) DNA B) tRNA C) rRNA D) mRNA E) tRNA, rRNA and mRNAs are all abundant

D). mRNA is the only type of RNA that is polyadenylated, and its poly A tail would be able to base-pair with the strands of poly T on the beads and thus stick to them. DNA would not be found in the sample, because the poly A tail is not encoded in the DNA and long runs of T are rare in DNA.

During transcription in __________________ cells, transcriptional regulators that bind to DNA thousands of nucleotides away from a gene's promoter can affect a gene's transcription. The __________________ is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcriptional start site. Transcriptional activators can also interact with histone __________________s, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone __________________s. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the __________________ found in interphase chromosomes.

During transcription in eucaryotic cells, transcriptional regulators that bind to DNA thousands of nucleotides away from a gene's promoter can affect a gene's transcription. The mediator is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcriptional start site. Transcriptional activators can also interact with histone acetylases, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone deacetylases. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the heterochromatin found in interphase chromosomes.

Which of the structures in Figure 3 is (are) not in a DNA chain? look at figure

E), uracil (C) and ribose (D) are found in RNA, but not normally found in DNA

Gene sequences correspond exactly to the respective protein sequences produced from them. If the statement is false, explain why it is false. A) True B) False

False. This statement is false for two reasons. First, genes often contain intron sequences. Second, genes always contain nucleotides flanking the protein-coding sequences that are required for the regulation of transcription and translation.

For a cell's genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________________. Various kinds of RNA are produced, each with different functions. __________________ molecules code for proteins, __________________ molecules act as adaptors for protein synthesis, __________________ molecules are integral components of the ribosome, and __________________ molecules are important in the splicing of RNA transcripts

For a cell's genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called transcription. Various kinds of RNA are produced, each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors for protein synthesis, rRNA molecules are integral components of the ribosome, and snRNA molecules are important in the splicing of RNA transcripts.

After treating cells with a mutagen, you isolate two mutants. One carries alanine and the other carries methionine at a site in the protein that normally contains valine. After treating these two mutants again with mutagen, you isolate mutants from each that now carry threonine at the site of the original valine (see Figure 2). Assuming that all mutations caused by the mutagen are due to single nucleotide changes, deduce the codons that are used for valine, alanine, methionine, and threonine at the affected site. (Refer to the codon table provided above).

Given that there are only single nucleotide changes, the only codons consistent with the changes are GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine.

Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and is therefore not expressed. An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression. Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.

In the absence of telomerase, the life-span of a cell and its progeny cells is limited. With each round of DNA replication, the length of telomeric DNA will shrink, until finally all the telomeric DNA has disappeared. Without telomeres capping the chromosome ends, the ends might be treated like breaks arising from DNA damage, or crucial genetic information might be lost. Cells whose DNA lacks telomeres will stop dividing or die. However, if telomerase is provided to cells, they may be able to divide indefinitely because their telomeres will remain a constant length despite repeated rounds of DNA replication

Cells can be very diverse; superficially, they come in various sizes, ranging from bacterial cells such Lactobacillus, which is few _____in ______.

Micrometers Millimeter

You have produced a monoclonal antibody that binds to the protein actin. To be sure that the antibody does not cross-react with other proteins, you test your antibody in a western blot assay on whole cell lysates that have been subjected to electrophoresis under nondenaturing conditions (shown in gel A) and denaturing conditions (shown in gel B). Does the antibody cross-react with other proteins? If so, does this explain the results in the two western blots? If not, how do you explain the differences observed? image looked

No, the antibody does not seem to cross-react with other proteins. In each western blot, there is only one band, indicating that only one protein is bound by the monoclonal antibody. Actin is a protein that forms long filaments. Under the nondenaturing conditions of the first gel, the filaments remain intact and, as a multiprotein complex, it migrates very slowly through the polyacrylamide matrix. In the case where SDS is added, actin filaments dissociate into monomers. Thus, the band is lower in panel B because the monomers have a lower molecular weight and migrated faster through the gel.

You are digesting a protein 625 amino acids long with the enzymes Factor Xa and thrombin, which are proteases that bind to and cut proteins at particular short sequences of amino acids. You know the amino acid sequence of the protein and so can draw a map of where Factor Xa and thrombin should cut it. You find, however, that treatment with each of these proteases for an hour results in only partial digestion of the protein, as summarized under the figure. List the segments (A-E) of the protein that are most likely to be folded into compact, stable domains. image. check

Segments B and D. To cut the protein chain, Factor Xa and thrombin must bind to their preferred cutting sites. If these sites are folded into the interior of a stable protein domain, it will be much more difficult for the proteases to gain access to them than if they are part of a relatively unstructured part of the chain. Hence, sites that are folded inside a protein domain are protected from cleavage by a protease. From the sizes of the fragments produced by digestion of the protein with Factor Xa, we can conclude that the enzyme does not cut at the sites in regions B or D, although it does cut in region E. From the sizes of the fragments produced by thrombin, we can conclude that this enzyme cuts at the sites in A, C, and E. Therefore, the segments of the protein that are most likely to be folded into compact stable domains are B and D.

Calculate how many different amino acid sequences there are for a polypeptide chain 10 amino acids long

The quantity 2010 ≅ 1013. This is a huge number.

Whereas ionic bonds form a(n) __________________, covalent bonds between atoms form a(n) __________________. These covalent bonds have a characteristic bond __________________ and become stronger and more rigid when two electrons are shared in a(n) __________________. Equal sharing of electrons yields a(n) __________________ covalent bond. If one atom participating in the bond has a stronger affinity for the electron, this produces a partial negative charge on one atom and a partial positive charge on the other. These __________________ covalent bonds should not be confused with the weaker __________________ bonds that are critical for the three-dimensional structure of biological molecules and for interactions between these molecules.

Whereas ionic bonds form a ((((salt))), covalent bonds between atoms form a ((((molecule.))) These covalent bonds have a characteristic bond ((((length)))) and become stronger and more rigid when two electrons are shared in a ((((double bond.)))) Equal sharing of electrons yields a ((((nonpolar))) covalent bond. If one atom participating in the bond has a stronger affinity for the electron, this produces a partial negative charge on one atom and a partial positive charge on the other. These polar covalent bonds should not be confused with the weaker noncovalent bonds that are critical for the three-dimensional structure of biological molecules and for interactions between these molecules.

Sometimes chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the cytosine in the sequence TCAT is deaminated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication? A) TTAT B) TUAT C) TGAT D) TAAT E) ATTA

a.)

The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process. Which choice below describes the known sequence of events in this process? A) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by repair proteins, the gap is filled by DNA polymerase, and the strand is sealed by DNA ligase. B) DNA repair polymerase simultaneously removes bases ahead of it and polymerizes the correct sequence behind it as it moves along the template. DNA ligase seals the nicks in the repaired strand. C) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by an exonuclease, and the gap is repaired by DNA ligase. D) A nick in the DNA is recognized, DNA repair proteins switch out the wrong base and insert the correct base, and DNA ligase seals the nick. E) RNA polymerase simultaneously removes bases ahead of it and polymerizes the correct sequence behind it as it moves along the template. DNA ligase seals the nicks in the repaired strand.

a.)

Given what you know about the differences between procaryotic cells and eucaryotic cells, rate the following things as "good" or "bad" processes to study in the model organism, E. coli. A) formation of the endoplasmic reticulum B) DNA replication C) how the actin cytoskeleton contributes to cell shape D) how cells decode their genetic instructions to make proteins E) how mitochondria get distributed to cells during cell division

a.) bad b.) good c.) bad d.) good e.) bad

Zebrafish (Danio rerio) are especially useful in the study of early development because their embryos ______________.

are transparent

Interphase chromosomes are about______ times less compact than mitotic chromosomes, but still are about______ times more compact than a DNA molecule in its extended form. A) 10; 1000 B) 20; 500 C) 5; 2000 D) 50; 200 E) 200; 50

b.)

Mitotic chromosomes were first visualized in the 1880s with the use of very simple tools: a basic light microscope and some dyes. Which of the following characteristics of mitotic chromosomes reflects how they were named? A) motion B) color C) shape D) location E) length

b.)

The N-terminal tail of histone H3 can be extensively modified, and depending on the number, location, and combination of these modifications, these changes may promote the formation of heterochromatin. What is the result of heterochromatin formation? A) increase in gene expression B) gene silencing C) recruitment of remodeling complexes D) displacement of histone H1 E) strengthens the nuclear envelope

b.)

The piece of RNA below includes the region that codes the binding site for the initiator tRNA needed in translation (t-met). Knowing that AUG codes for methionine (the start of translation), which amino acid will be on the tRNA that is the first to bind to the A-site of the ribosome? 5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′ A) methionine B) arginine C) cystine D) valine E) leucine

b.)

You have a piece of DNA that includes the following sequence: 5′-ATAGGCATTCGATCCGGATAGCAT-3′ 3′-TATCCGTAAGCTAGGCCTATCGTA-5′ Which of the following RNA molecules could be transcribed from this piece of DNA? A) 5′-UAUCCGUAAGCUAGGCCUAUGCUA-3′ B) 5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′ C) 5′-UACGAUAGGCCUAGCUUACGGAUA-3′ D) none of the above E) all of the above (A, B, and C)

b.)

The first task you are assigned in your summer laboratory job is to prepare a concentrated NaOH stock solution. The molecular weight of NaOH is 40. How many grams of solid NaOH will you need to weigh out to obtain a 500 ml solution that has a concentration of 10 M? (a) 800 g (b) 200 g (c) 400 g (d) 160 g

b.) 200 g

) Nucleosomes are formed when DNA wraps _____ times around the histone octamer in a ______ coil. A) 2.0; right-handed B) 2.5; left-handed C) 1.7; left-handed D) 1.3; right-handed E) 5.2; left-handed

c.)

) You are interested in understanding the gene regulation of Lkp1, a protein that is normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1 gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene, place it upstream of the gene for green fluorescent protein (GFP), and insert this entire piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells but not in heart cells, an expression pattern similar to the normal expression of the LKP1 gene. Further experiments demonstrate that there are three regions in the promoter, labeled A, B, and C in the figure below, that contribute to this expression pattern. Assume that a single and unique transcription factor binds each site such that protein X binds site A, protein Y binds site B, and protein Z binds site C. You want to determine which region is responsible for tissue-specific expression, and create mutations in the promoter to determine the function of each of these regions. In the figure below, if the site is missing, it is mutated such that it cannot bind its corresponding transcription factor. Which of the following proteins are likely to act as gene repressors? A) factor X B) factor Y C) factor Z D) none of the above E) there is not sufficient information to identify the likely repressor(s)

c.)

Below is a segment of RNA from the middle of an mRNA. 5′-UAGUCUAGGCACUGA-3′ If you were told that this segment of RNA was part of the coding region of an mRNA for a large protein, how many amino acids could accurately deduce? Refer to the genetic code chart for assistance. A) 2 B) 3 C) 4 D) 5 E) 6

c.)

In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes. During which process does swapping occur? A) DNA replication B) DNA repair C) meiosis D) transposition E) cytokinesis

c.)

The biosynthetic pathway for the two amino acids E and H is shown schematically. You are able to show that E inhibits enzyme V, and H inhibits enzyme X. Enzyme T is most likely to be subject to feedback inhibition by __________________ alone. image

c.)

How does the total number of replication origins in bacterial cells compare with the number of origins in human cells? A) 1 versus 100 B) 5 versus 500 C) 10 versus 1000 D) 1 versus 10,000 E) This number is not known for prokaryotic cells

d.)

Lysozyme is an enzyme that specifically recognizes bacterial polysaccharides, which renders it an effective antibacterial agent. Into what classification of enzymes does lysozyme fall? A) isomerase B) protease C) nuclease D) hydrolase E) glycosylase

d.)

The human genome has enough DNA to stretch more than 2 m. However, this DNA is not contained in a single molecule; it is divided into linear segments and packaged into structures called chromosomes. What is the total number of chromosomes found in each of the somatic cells in your body? A) 22 B) 23 C) 44 D) 46 E) 48

d.)

You are interested in examining the Psf gene. It is known that Psf is normally produced when cells are exposed to high levels of both calcium (Ca2+) and magnesium (Mg2+). MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are involved in regulating its transcription. MetA binds to the "A" site in the promoter region, MetB to the "B" site, and MetC to the "C" site. You create binding site mutations in the A, B, and C sites and observe what happens to transcription of the Psf gene. Your results are summarized in the Table below. For this table: -, no transcription of Psf; +, low level of transcription of Psf; ++, high levels of transcription of Psf. Which of the following proteins are likely to act as gene activators? A) MetA only B) MetB only C) MetC only D) Both MetA and MetC E) Both MetA and MetB

d.)

Which of the following chemical groups is not used to construct a DNA molecule? (two answers) A) five-carbon sugar B) phosphate C) nitrogen-containing base D) six-carbon sugar E) cyclic AMP

d.) e.)

You have a concentrated stock solution of 10 M NaOH and want to use it to produce a 150 ml solution of 3 M NaOH. What volume of water and stock solutions will you measure out to make this new solution? (a) 135 ml of water, 15 ml of NaOH stock (b) 115 ml of water, 35 ml of NaOH stock (c) 100 ml of water, 50 ml of NaOH stock (d) 105 ml of water, 45 ml of NaOH stock

d.) 105 ml of water, 45 ml of NaOH stock

. The pH of an aqueous solution is an indication of the concentration of available protons. However, you should not expect to find lone protons in solution; rather, the proton is added to a water molecule to form a(n) ______________ ion. (a) hydroxide (b) ammonium (c) chloride (d) hydronium

d.) hydronium

8) Which protein(s) relate to continuous replication. A) primase B) single-strand binding protein C) sliding clamp D) RNA primers E) leading strand

e.)

The human genome encodes about 24,000 genes. Approximately how many genes does the typical differentiated human cell express at any one time? A) 24,000—all of them B) between 21,500 and 24,000—at least 90% of the genes C) between 5 and 15 D) less than 2500 E) between 5000 and 15,000

e.)

An element may have more than a one isotope. Isotopes have different atomic weights, but exhibit the same chemical behavior. Carbon-14 is an unstable isotope of carbon that decays very slowly. Compared to the common, stable carbon-12 isotope, carbon-14 has two additional ______________.

neutrons

C. elegans is a nematode. During its development, it produces more than 1000 cells However, the adult worm only has 959 somatic cells. The process by which 131 cells are specifically targeted for destruction is called

programmed cell death

Fluorescence

requires the use of two sets of filters. The first filter narrows the wavelength range that reaches the specimen an the second blocks out all wavelength that pass back up to the eyepiece except for those emitted by the dye in the sample.

Confocal

scans the specimen with a focused laser beam to obtain a series of twodimensional optical sections, which can be used to reconstruct an image of the specimen in three-dimensions. The laser excites a fluorescent dye molecule, and the emitted light from each illuminated point is captured through a pinhole and recorded by a detector