ESSENTIAL CELL BIOLOGY : ALL CHAPTERS COVERED

11. Larger molecules have hydrogen-bonding networks that contribute to specific, high- affinity binding. Smaller molecules such as urea can also form these networks. How many hydrogen bonds can urea (see figure) form if dissolved in water? a) 6 (b) 5 (c) 3 (d) 4 ** SEE FIGURE**

(a). Urea can form at least six hydrogen bonds in water: two from the oxygen atom and one from each hydrogen atom.

By definition, procaryotic cells do not possess __________. A) a nucleus B) replication machinery C) ribosomes D) membrane bilayers E) DNA polymerase

A

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. A) insect B) bird C) amphibian D) mammal E) prokaryotic organism

A

11) 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)

3) 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

14) 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

15) 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

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. 10) 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)

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) 4

1) You have purified DNA from your recently deceased goldfish. Which of the following restriction nucleases would you use if you wanted to end up with DNA fragments with an average size of 70 kilobase pairs (kb) after complete digestion of the DNA? The recognition sequence for each enzyme is indicated in the right-hand column. A) Sau3AI GATC B) BamHI GGATCC C) NotI GCGGCCGC D) XzaI GAAGGATCCTTC

C). A restriction nuclease that has a 4-base-pair recognition sequence cuts on average once every 44 or 256 base pairs; one that has a 6-base-pair recognition sequence cuts once every 46 or 4096 base pairs; one that has an 8-base-pair recognition sequence cuts once every 48 or 65,536 base pairs; one that has a 12- base-pair recognition sequence cuts once every 412 or 16 million base pairs. Thus, to obtain fragments of about 70 kb in size, you would cut with a nuclease that recognizes an 8-base-pair site

2) 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) and E)

10) 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) hydrolase

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

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 _____________. A) it can withstand extremely cold climates B) it can reproduce in 8-10 weeks C) it produces thousands of offspring per plant D) it does not undergo programmed cell death E) Both B) and C) are true

E

Which of the structures in Figure 3 is (are) not in a DNA chain? A) (A) B) (B) C) (C) D) (A) and (D) E) (C) and (D)

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

To meet a challenge or develop a new function, evolution essentially builds from first principles, designing from scratch, to find the best possible solution. A) True B) False

False

The increased complexity of humans compared with flies and worms is largely due to the vastly larger number of genes in humans. A) True B) False

False The number of genes differs only by about a factor of two. It is thought that the increased complexity of humans is due largely to differences in when and where the genes are expressed.

7) 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.

4) 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)..]

5) 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.

You are studying a protein that contains the peptide sequence RDWKLVI. The part of the DNA encoding this peptide is included in the sequence shown below. 5′-GGCGTGACTGGAAGCTAGTCATC-3′ 3′-CCGCACTGACCTTCGATCAGTAG-5′ This sequence does not contain any HindIII restriction enzyme sites; the target sequence for the HindIII restriction nuclease is shown in Figure 9. Figure 9 Your goal is to create a HindIII site on this plasmid without changing the coding sequence of the protein. Explain how you would do this. (Refer to the codon table above.)

It is unlikely that child would have the disease, because it is unlikely that the mutation is carried in the germ line. Statistically, it is probable the mutation occurred in a somatic cell and not germ cell as there are many more somatic cells. Only mutations in germ cells are passed on to progeny.

For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or phrases will be used; use each word or phrase only once. Sexual reproduction in a multicellular organism involves specialized reproductive cells, called __________________s, which come together to form a __________________ that will divide to produce both reproductive and __________________ cells. A point mutation in the DNA is considered a __________________ mutation if it changes a nucleotide that leads to no phenotypic consequence; a point mutation is considered __________________ if it changes a nucleotide within a gene and causes the protein to be non-functional. common somatic gamete neutral homologous intron deleterious cellulose unequal zygote

Sexual reproduction in a multicellular organism involves specialized reproductive cells, called ___gametes__________s, which come together to form a ____zygote______________ that will divide to produce both reproductive and ____somatic______________ cells. A point mutation in the DNA is considered a _______neutral___________ mutation if it changes a nucleotide that leads to no phenotypic consequence; a point mutation is considered ____deleterious___________ if it changes a nucleotide

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.

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.

13) Which of the restriction nucleases listed below can potentially cleave a segment of cDNA that encodes the peptide KIGDACF? Use the Genetic Code Table to help determine the correct answer. **SEE FIGURE** Restriction Nuclease EcoRI HindIII NSI Recognition Sequence GAATTC AAGCTT ATGCAT Assume that you are trying to get the larger portion of plasmid to reassemble into a circle.

The nucleotide sequences that can encode the peptide KIGDACF are shown below. The enzyme NsiI cleaves at ATGCAT. **SEE FIGURE**

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.

12) 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.

13) 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.

Proteins required for growth, metabolism, and cell division are more highly conserved than those involved in development and in response to the environment. A) True B) False

True. All organisms need to perform a similar basic set of fundamental functions, such as those for metabolism, protein synthesis, and DNA replication. Proteins involved in these functions are shared by descent, and their evolution is constrained. Different species and cells are likely to require different developmental paths and to encounter different environmental challenges, so the proteins involved in these processes will tend to be more variable. For example, bacteria do not undergo elaborate developmental programs and so lack many of the regulators of development found in eucaryotes.

Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial cells such as Lactobacillus, which is a few __________ in length, to larger cells such as a frog's egg, which has a diameter of about one ___________. Despite the diversity, cells resemble each other to an astonishing degree in their chemistry. For example, the same 20 _____________ are used to make proteins. Similarly, the genetic information of all cells is stored in their __________. 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. amino acids micrometer(s) viruses DNA millimeter(s) yeast fatty acids plants meter plasma membranes

micrometers millimeter amino acids DNA viruses

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.

6) 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′.

12) 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

16) 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

1) 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.

5) You are interested in a single-stranded DNA molecule that contains the following sequence: 5′- ..... GATTGCAT .... -3′ Which molecule can be used as a probe that will hybridize to your sequence of interest? A) 5′-GATTGCAT-3′ B) 5′-TACGTTAG-3′ C) 5′-CTAACGTA-3′ D) 5′-ATGCAATC-3′

D

6) 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

Some types of gene are more highly conserved than others. Which gene function is more likely to be highly conserved in the global genome (e.g., humans, yeast, bacteria). hormone production lipid synthesis

Lipid synthesis is fundamental to the growth and proliferation of all cells, including bacteria, and thus is likely to be highly conserved from species to species.

6) 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.

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

between 5000 and 15,000

3) 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

see figure labels

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

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.

8) Name three features that a cloning vector for use in bacteria must contain. Explain your answers. 1. 2. 3.

A cloning vector for use in bacteria must contain the following: 1. a bacterial replication origin (to allow the plasmid to be replicated) 2. at least one unique restriction site (to allow easy insertion of foreign DNA) 3. an antibiotic-resistance gene or some other selectable marker gene (to allow selection for bacteria that have taken up the recombinant plasmids)

15. Eucaryotic cells have their DNA molecules inside their nuclei. However, to package the DNA all into such a small volume requires the cell to use specialized proteins called histones. Histones have amino acid sequences enriched for lysines and arginines. Lysine side chains are substrates for enzymes called acetylases. A diagram of an acetylated lysine side chain is shown. How do you think the acetylation of lysines in histone proteins will affect the ability of a histone to perform its role? ** SEE FIGURE**

A histone with acetylated lysine residues will not be as good at packaging the DNA. The addition of the acetyl group to the terminal amino on the lysine side chain lowers the histone's net positive charge, which makes it less effective at buffering the negative charges on the DNA backbone.]

6) For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. A nuclease hydrolyzes the __________________ bonds in a nucleic acid. Nucleases that cut DNA only at specific short sequences are known as __________________. A) phosphodiester B) endonucleases C) exonucleases D) restriction nucleases E) hydrogen

A nuclease hydrolyzes the phosphodiester bonds in a nucleic acid. Nucleases that cut DNA only at specific short sequences are known as restriction nucleases. DNA composed of sequences from different sources is known as recombinant DNA. Gel electrophoresis can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro by the polymerase chain reaction technique.

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) 5′-CCA-3′

4) Figure 3 shows a restriction map of a piece of DNA containing your favorite gene. The arrow indicates the position and orientation of the gene in the DNA. In part B) of the figure are enlargements showing the portions of the DNA whose sequences have been used to make oligonucleotide probes A, B, C, and D. Which of the oligonucleotides can be used to detect the gene in each of the following? Figure 3 A) A Southern blot of genomic DNA cut with HindIII. B) A Northern blot.

A) All four oligonucleotide probes. B) Oligonucleotide probe B and oligonucleotide probe D. Both the upper and lower strands of DNA are present in genomic Southern blots, so all four oligonucleotides will hybridize to either Southern blot. (Oligonucleotides A and B will still be able to hybridize to genomic DNA cut with BglII, because they can still base-pair with the individual fragments that result from the digest.) Northern blots contain only RNA, which has the sequence of the upper strand of the DNA. Hence, only oligonucleotide probe B and oligonucleotide probe D will hybridize to a Northern blot.

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.

7) 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.

3) You have accidentally torn the labels off two tubes, each containing a different plasmid, and now do not know which plasmid is in which tube. Fortunately, you have restriction maps for both plasmids, shown in Figure 2. You have the opportunity to test just one sample from one of your tubes. You have equipment for agarose-gel electrophoresis, a standard set of DNA size markers, and the necessary restriction enzymes. A) Outline briefly the experiment you would do to determine which plasmid is in which tube. B) Which restriction enzyme or combination of restriction enzymes would you use in this experiment?

A) You would first digest your sample with a combination of restriction enzymes selected so that they give a set of fragment sizes that could have come from only one of the plasmids. Then you would run the resulting mixture of DNA fragments on a gel alongside a set of size markers and determine the size of each fragment. By looking at the restriction maps, you should then be able to match your results to one of the plasmids. B) Digestion with any of the following combinations will enable you to distinguish which plasmid you have: HindIII + BglII; EcoRI + BglII; EcoRI + BglII + HindIII. The plasmids are the same size, so you cannot distinguish between them simply by making a single cut (with HindIII) and determining the size of the complete DNA by gel electrophoresis. Nor can you distinguish them by cutting with all four restriction nucleases, Because the set of fragment sizes produced from both plasmids will be the same. Cutting with BamHI or EcoRI on their own is not sufficient because you will get bands of the same size from both plasmid A and plasmid B. The only difference between the two plasmids is in the location of the BglII site relative to the two BamHI sites, so if you cut with an enzyme that cuts outside the BamHI fragment and with BglII, you will get different-sized fragments from the two plasmids.

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 inst ructions to make proteins E)how mitochondria get distributed to cells during cell division

A) bad B) good C) bad D) good E) bad

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) lysine 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′.

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. 9) 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)

11) You want to design a DNA probe used for hybridization to isolate a clone from a cDNA library. Which of the following statements about DNA probes is true? A) The shorter the DNA probe used to probe the library, the greater the number of colonies to which the probe will hybridize. B) A DNA probe that contains sequences that span two exons is better suited to the purpose than a DNA probe that only contains sequences from one exon. C) A DNA probe that contains sequences immediately upstream of the DNA that codes for the first methionine in the open reading frame will usually not hybridize to clones in a cDNA library. D) Hybridization of a DNA probe to the plasmid of interest will permit the detection of the clone of interest; labeling of the DNA probe is not necessary. E) All (A-D) are true

A). The shorter the DNA probe, the more likely it is that that particular sequence will show up in the genome at random. cDNA libraries contain sequences represented by exons, so it does not really matter whether or not the probe spans two exons (choice B)). mRNAs usually have 5′ untranslated regions that should be represented in a cDNA library, so choice C) is not true. DNA probes are usually labeled (for example, with radioactivity) for visualization (choice D)).

20)Your friend is working in a lab to study how yeast cells adapt to growth on different carbon sources. He grew half of his cells in the presence of glucose and the other half in the presence of galactose. Then he harvested the cells and isolated their DNA with a gentle procedure that leaves nucleosomes and some higher-order chromatin structures intact. He treated the DNA briefly with a low concentration of M-nuclease, a special enzyme that easily degrades protein-free stretches of DNA. After removing all the proteins, he separated the resulting DNA on the basis of length. Finally, he used a procedure to visualize only those DNA fragments from a region near a particular gene called Sweetie or another gene called Salty. The separated DNA fragments are shown in the gel below. Each vertical column, called a lane, is from a different sample. DNA spots near the top of the figure represent DNA molecules that are longer than those near the bottom. Darker spots contain more DNA than fainter spots. The lanes are as follows: 1. "marker" containing known DNA fragments of indicated lengths 2. cells grown in glucose, DNA visualized near Sweetie gene 3. cells grown in galactose, DNA visualized near Sweetie gene 4. cells grown in glucose, DNA visualized near Salty gene 5. cells grown in galactose, DNA visualized near Salty gene **look at figure** A. The lowest spot (as observed in lanes 2, 4, and 5) has a length of about 150 nucleotides. Can you propose what it is and how it arose? B. What are the spots with longer lengths? Why is there a ladder of spots? C. Notice the faint spots and extensive smearing in lane 3, suggesting the DNA could be cut almost anywhere near the Sweetie gene after growth of the cells in galactose. This was not observed in the other lanes. What probably happened to the DNA to change the pattern between lanes 2 and 3? D. What kinds of enzyme might have been involved in changing the chromatin structure between lanes 2 lane 3? E. Do you think that gene expression of Sweetie is higher, lower, or the same in galactose compared to glucose? What about Salty?

A. The lowest spot represents DNA of a length similar to that of the segment of DNA found in a nucleosome core particle. Partial digestion with an enzyme such as Mnuclease causes breaks in the DNA backbone primarily within the linker DNA or other DNA segments not bound tightly to histones. Thus, this band probably comprises the DNA bound tightly to a single histone octamer and it arose by cutting the linker DNA outside a single nucleosome core particle. B. The ladder of bands with longer lengths probably corresponds to stretches of DNA associated with increasing numbers of nucleosomes (1, 2, 3, 4, 5, and so on). In support of this proposal, adjacent bands differ in size by roughly 200 nucleotides, which is the length of DNA found in a nucleosome core particle plus neighboring linker DNA. This interpretation must mean that the M-nuclease digestion did not go to completion, because if all non-nucleosomal DNA were digested completely, the samples would contain only the 150-base-pair fragment. C. Given the ability of M-nuclease to cut anywhere near Sweetie after growth in galactose, it seems that the DNA is no longer protected from digestion by binding to histones. Perhaps the wrapping of DNA within the nucleosomes has been loosened considerably. This change in the nucleosomes must be specific to the Sweetie gene, because it is not seen at the Salty gene or throughout the genome. D. The main candidates for enzymes that catalyzed the nucleosome alterations near Sweetie are chromatin-remodeling complexes and enzymes that covalently modify histone tails with methyl, acetyl, or phosphate groups. E. As the chromatin seems to have been loosened near Sweetie, it seems likely that Sweetie gene expression is increased when cells are grown in galactose rather than glucose, whereas Salty gene expression is likely to be the same under the two conditions. Perhaps the Sweetie gene contains instructions for a protein that is required for cells to metabolize galactose but not glucose.

13. 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..

12. The amino acid histidine is often found in enzymes. Depending on the pH of its environment, sometimes histidine is neutral and at other times it acquires a proton and becomes positively charged. Consider an enzyme with a histidine side chain that is known to have an important role in the function of the enzyme. It is not clear whether this histidine is required in its protonated or its unprotonated state. To answer this question you measure enzyme activity over a range of pH, with the results shown in the graph. Which form of histidine is necessary for the active enzyme? ** SEE FIGURE**

Assuming that the change in enzyme activity is due to the change in the protonation state of histidine, the enzyme must require histidine in the protonated, charged state. The enzyme is active only at low, acidic, pH, where the proton (or hydronium ion) concentration is high; thus, the loss of a proton from histidine will be disfavored so that histidine is likely to be protonated.

8) 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

2) You have a piece of circular DNA that can be cut by the restriction nucleases XhoI and SmaI, as indicated in Figure 1. If you were to cut this circular piece of DNA with both XhoI and SmaI, how many fragments of DNA would you end up with? A) 1 B) 2 C) 3 D) 4 E) 5

B

8) 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 A) directed cell pruning. B) programmed cell death. C) autophagy. D) necrosis. E) surgical strike.

B

10) Figure 5 shows the recognition sequences and sites of cleavage for the restriction enzymes SalI, XhoI, PstI, and SmaI, and also a plasmid with the sites of cleavage for these enzymes marked. SEE FIGURE 5 In which of the treatments can the plasmid re-form into a circle by the addition of DNA ligase after treating the cut DNA with DNA polymerase in a mixture containing the four deoxyribonucleotides?

B) 1, 2, and 4. SmaI cuts and leaves a blunt end. Addition of DNA polymerase and the four deoxynucleotides will fill in the 5′ overhangs generated by digestion with SalI and XhoI, leaving blunt ends. DNA ligase joins the blunt ends. However, 3′ overhangs (that is, those generated by PstI) will not be filled in, because DNA polymerase moves in a 5′ to 3′ direction. DNA ligase will not join 3′ overhangs to blunt-ended DNA, which are the situations presented in treatments 3 and 5.

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) 5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′

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

B) False

15) 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

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

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) The genetic code is redundant.

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) arginine

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) arginine

7) When a mutation arises, the three possible consequences in order from most likely to least likely are: A) beneficial detrimental selectively neutral B) selectively neutral detrimental beneficial C) selectively neutral beneficial detrimental D) detrimental selectively neutral beneficial E) beneficial selectively neutral detrimental

B) selectively neutral detrimental beneficial The order is selectively neutral, detrimental, and beneficial. Most nucleotide changes in the genome, or mutations, will have little or no effect on the fitness of the individual because many changes are not located in regions that encode a protein or regulate the expression of a gene. Even changes within a coding region may not change the amino acid encoded or may cause a conservative amino acid change, for example from one small nonpolar amino acid to another. Most changes that have a functional consequence will interfere with the regulation of a gene or the behavior of the encoded protein, usually rendering it useless and occasionally making it harmful or yielding a new function. Only very rarely will a mutation improve the performance of the gene or its encoded protein.

9) Figure 4 shows the cleavage site of several restriction nucleases. SEE Figure 4 You cut a vector using the PclI restriction nuclease. Which of the following restriction nucleases will generate a fragment that can be ligated into this cut vector with the addition of only ligase and ATP? A) HindIII B) NcoI C) MmeI D) NspV E) all of the restriction nucleases

B). However, you will not be able to excise the fragment after ligation, because you will destroy both the PclI site and the NcoI site, creating a new site with the sequence 5′-ACATGG-3′ 3′-TGTACC-5′

12) You have the amino acid sequence of a protein and wish to search for the gene encoding this protein in a DNA library. Using this protein sequence, you deduce a particular DNA sequence that can encode this protein. Why is it unwise to use only this DNA sequence you have deduced as the probe for isolating the gene encoding your protein of interest from the DNA library?

Because most amino acids can be encoded by more than one codon, a given sequence of amino acids could be encoded by several different nucleotide sequences. Probes corresponding to all these possible sequences have to be synthesized, to be sure of including the one that corresponds to the actual nucleotide sequence of the gene and thus will hybridize with it

17) 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

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) Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold.

5) Which of the following changes is least likely to arise from a point mutation in a regulatory region of a gene? A) a mutation that changes the time in an organism's life during which a protein is expressed B) a mutation that eliminates the production of a protein in a specific cell type C) a mutation that changes the subcellular localization of a protein D) a mutation that increases the level of protein production in a cell E) A-D are correct

C) a mutation that changes the subcellular localization of a protein Information for the subcellular localization of a protein is usually encoded within the translated portion of the gene. A, B, and D would be essentially mutations in regulatory regions.

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) factor Z

Alternative exons can arise through the duplication and divergence of existing exons. What type of mutation below would be least tolerated during the evolution of a new exon? A) a nucleotide change of A to G B) a deletion of three consecutive bases. C) mutation of the first nucleotide in the intron D) a nucleotide change that alters a TT dinucleotide to AA E) a point mutation that changes alanine to glycine

C) mutation of the first nucleotide in the intron The first two nucleotides in the intron are critical for signaling the exon-intron boundary; changing them would make the exon unable to be properly spliced.

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) up to 16

2) 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). A hydrogen bond forms between a polar side chain and an aromatic side chain. 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.

14) 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.

4) 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.

4) 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.

14. 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

D) All amino acids contain an NH2 and a COOH group 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.

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 is the template strand; RNA polymerase moves along this strand from 3′ to 5′.

6) Which of the following would contribute most to successful exon shuffling? A) shorter introns B) a haploid genome C) exons that code for more than one protein domain D) introns that contain regions of similarity to one another E) introns that do not encode alternative slice sites

D) introns that contain regions of similarity to one another Exon shuffling is facilitated by long introns (thus choice A) is incorrect) and by short exons that each code for one protein domain (thus choice C) is incorrect). Because exon shuffling can occur by recombination between introns, introns with regions of similarity to one another will facilitate shuffling. A haploid genome will probably be less prone to exon shuffling than a diploid genome (thus choice B) is incorrect), because having two copies of each gene allows an organism to keep one copy of the gene as a backup while it shuffles the other copy.

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

D) nucleotide polymerization occurs only in the 5′-to-3′ direction

3) 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.

For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or phrases will be used; use each word or phrase only once. 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 __acetlases__________, which alter chromatin by modifying lysines in the tail of histone proteins to allow greateraccessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone __deactylases____. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the ___heterochromatin_______________ found in interphase chromosomes.

This statement is for questions 8, 9, and 10. Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. 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

Combinatorial control of gene expression __________________________. A) involves every gene using a different combination of transcriptional regulators for its proper expression B) involves every gene using a the same identical combination of transcriptional regulators for its proper expression C) involves only the use of gene activators used together to regulate genes appropriately D) is seen only when genes are arranged in operons E) involves groups of transcriptional regulators working together to determine the expression of a gene

E) involves groups of transcriptional regulators working together to determine the expression of a gene

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.

For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. 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. incorporation rRNA transmembrane mRNA snRNA tRNA pRNA transcription proteins 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 __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.

5) 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.

7) 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? **see figure**

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.

5) 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. ** See figure**

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.

The genomes of some vertebrates are much smaller than those of others. For example, the genome of the puffer fish Fugu is much smaller than the human genome, and even much smaller than those of other fish, primarily because of the small size of its introns. Describe a mechanism that might drive evolution toward small introns or loss of introns and could therefore account for the evolutionary loss of introns according to the "introns early" hypothesis.

Spontaneous deletions or selection pressure to decrease the time or cost of DNA replication may cause a loss of introns.

Consider a gene with a particular function. Mutation X and mutation Y each cause defects in the function of the encoded protein, yet a gene containing both mutations X and Y encodes a protein that works even better than the original protein. The odds are exceedingly small that a single mutational event will generate both mutations X and Y. Explain a simple way that an organism with a mutant gene containing both mutations X and Y could arise during evolution.

The simplest way to evolve the new gene is by duplication and divergence. If the gene is duplicated, then the cell or lineage can maintain one functional, intact old copy of the original gene and can thus tolerate the disabling mutations in the other copy. The other copy can first be modified by the X or Y mutation that impairs its function; second, at some later time, the gene with the single mutation canacquire the additional mutation to yield the doubly mutant X + Y gene with the new or improved function.

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.

7) For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. Two fragments of DNA can be joined together by __________________. Enzymes that cut DNA straight across the double helix produce fragments of DNA with ___________. A) blunt ends B) DNA polymerase C) staggered ends D) DNA ligase E) vectors

Two fragments of DNA can be joined together by DNA ligase. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with blunt ends. A fragment of DNA is inserted into a vector in order to be cloned in bacteria. A cDNA library contains a collection of DNA clones derived from mRNAs. A genomic library contains a collection of DNA clones derived from chromosomal DNA.

Your friend has just returned from a deep sea mission and claims to have found a new single - celled life form. He believes this new life form may not have descended from the common ancestor that all types of life on Earth share. You are convinced that he must be wrong, and you manage to extract DNA from the cells he has discovered. He says that the mere presence of DNA is not enough to prove the point: his cells might have adopted DNA as a useful molecule quite independently of all other known life forms. What would you do to persuade him that he was wrong?

You could use modern technology to discover the sequence of the DNA. If you are right, and he is wrong, you would expect to find parts of this sequence that are unmistakably similar to corresponding sequences in other, familiar, living organisms; it would be highly improbable that such similar sequences would have evolved independently. You could, of course, also analyze other feature s of the chemistry of his cells; for example, do they contain proteins made of the same set of 20 amino acids? (But he would probably not find such general chemical similarities as persuasive as detailed similarities of DNA sequence)

You fertilize egg cells from a healthy plant with pollen (which contains the male germ cells) that has been treated with DNA - damaging agents. You find that some of the offspring have defective chloroplasts, and that this characteristic can be passed on to future generations. This surprises you at first because you happen to know that the male germ cell in the pollen grain contributes no chloroplasts to the fertilized egg cell and thus to the offspring. What can you deduce from these results

Your results show that not all of the information required for making a chloroplast is encoded in the chloroplast's own DNA; some, at least, must be encoded in the DNA carried in the nucleus. The reasoning is as follows. Genetic information is carried only in DNA, so the defect in the chloroplasts must be due to a mutation in DNA. But all of the chloroplasts in the offspring (and thus all of the chloroplast DNA) must derive from those in the female egg cell, since chloroplasts only arise from other chloroplasts. Hence, all of the chloroplasts contain undamaged DNA from the female parent's chloroplasts. In all the cells of the offspring, however, half of the nuclear DNA will have come from the male germ cell nucleus, which combined with the female egg nucleus at fertilization. Since this DNA has been treated with DNA - damaging agents, it must be the source of the heritable chloroplast defect. Thus, some of the information required for making a chloroplast is encoded by the nuclear DNA.