Celluspurningar

5-5. (S/Ó) DNA repair mechanisms all depend on the existence of two copies of the genetic information, one in each of the two homologous chromosomes.

False. Repair of damage to a single strand by base excision repair or nucleotide excision repair, for example, depends on just the two copies of genetic information contained in the two strands of the DNA double helix. By contrast, precise repair of damage to both strands of a duplex—a double-strand break, for example—requires information from a second duplex, either a sister chromatid or a homolog.

5-13. With age, somatic cells are thought to accumulate genomic "scars" as a result of the inaccurate repair of double-strand breaks by nonhomologous end joining (NHEJ). Estimates based on the frequency of breaks in primary human broblasts suggest that by age 70, each human somatic cell may carry some 2000 NHEJ-induced mutations due to inaccurate repair. If these mutations were distributed randomly around the genome, how many protein-coding genes would you expect to be affected? Would you expect cell function to be compromised? Why or why not? (Assume that 2% of the genome—1.5% protein-coding and 0.5% regulatory—is crucial information.)

If the inaccurately repaired breaks were randomly distributed around the genome, then 2% of them would be expected to alter crucial coding or regulatory information. us, the functions of about 40 genes (0.02 × 2000) would be compromised in each cell, although the speci c genes would vary from cell to cell. Because not all genes are expressed in every cell, gene mutations in some cells would be without consequence. In addition, because the human genome is diploid, the e ect on cell func- tion of mutations in expressed genes would be mitigated by the remain- ing allele. For most loci, one functional allele (50% of normal protein) is adequate for normal cell function; however, for some loci, 50% is not adequate. us, the mutations would be expected to compromise the functions of some cells.

4-6. DNA isolated from the bacterial virus M13 contains 25% A, 33% T, 22% C, and 20% G. Do these results strike you as peculiar? Why or why not? How might you explain these values?

In ALL samples of double-stranded DNA, the numbers of As and Ts (hence their percentages) are equal since they always pair with each other. The same is true for G and C. Results such as this one stood out as odd in the days before the structure of DNA was known. Now it is clear that, while all cellular DNA is double stranded, certain viruses contain single-stranded DNA. The genomic DNA of the M13 virus, for example, is single stranded. In single-stranded DNA, A is not paired with T, nor G with C, and so the A = T and C = G rules do not apply.

4-3. (S/Ó) Nucleosomes bind DNA so tightly that they cannot move from the positions where they are first assembled.

False. By using the energy of ATP hydrolysis, chromatin remodeling complexes can catalyze the movement of nucleosomes along DNA, or even dissociate a nucleosome completely from the DNA.

4-12. Why is a chromosome with two centromeres (a dicentric chromosome) unstable? Would a backup centromere not be a good thing for a chromosome, giving it two chances to form a kinetochore and attach to microtubules during mitosis? Would that not help to ensure that the chromosome did not get left behind at mitosis?

A dicentric chromosome is unstable because the two kinetochores have the potential to interfere with one another. Normally, microtubules from the two poles of the spindle apparatus attach to opposite faces of a single kinetochore in order to separate the individual chromatids at mitosis. If a chromosome contains two centromeres, half of the time the microtubules from one of the poles will attach to the two kinetochores associated with one chromatid, while the microtubules from the other pole will attach to the two kinetochores associated with the other chromatid. Division can then occur satisfactorily. The other half of the time, the microtubules from each pole will attach to kinetochores that are associated with different chromatids. When that happens, each chromatid will be pulled to opposite spindle poles with enough force to snap it in two. Thus, two centromeres are bad for a chromosome, causing chromosome breaks— rendering it unstable.

5-15. In addition to correcting DNA mismatches, the mismatch repair system functions to prevent homologous recombination from taking place between similar but not Problems p5.39/5.27/Q5.2/Q5.3 identical sequences. Why would recombination between similar, but nonidentical sequences pose a problem for human cells?

A large percentage of the human genome is made up of repetitive elements such as Alu sequences, which are scattered among the chromosomes. If, for example, recombination were to occur between two such sequences that were on di erent chromosomes, a translocation would be generated. Unrestricted recombination between such repeated elements would quickly rearrange the genome beyond recognition. Di erent rearrangements in di erent individuals would lead to large numbers of nonviable progeny, putting the species at risk. This calamity is avoided through the action of the mismatch repair system. Repeated sequences around the genome di er by a few percent of their sequence. When recombination intermediates form between them, many mismatches are present in the heteroduplex regions. When the mismatch repair system detects too high a frequency of mismatches, it aborts the recombination process in some way. is surveillance mechanism ensures that sequences that successfully recombine are nearly identical, as expected for sequences at the same locus on homologous chromosomes.

5-12. If you compare the frequency of the sixteen possible dinucleotide sequences in the E. coli and human genomes, there are no striking differences except for one dinucleotide, 5ʹ-CG-3ʹ. The frequency of CG dinucleotides in the human genome is significantly lower than in E. coli and significantly lower than expected by chance. Why do you suppose that CG dinucleotides are underrepresented in the human genome?

At many sites in vertebrate cells, the sequence 5ʹ-CG-3ʹ is selectively methylated on the cytosine base. Spontaneous deamination of methyl- C produces T. A special DNA glycosylase recognizes a mismatched base pair involving T in the sequence TG, and removes the T. is DNA repair mechanism is clearly not 100% e ective, as methylated C nucleotides are common sites for mutation in vertebrate DNA. Over time, the enhanced mutation rate of CG dinucleotides has led to their preferential loss, accounting for their underrepresentation in the human genome.

4-8. Human DNA contains 20% C on a molar basis. What are the mole percents of A, G, and T?

Because C always pairs with G in duplex DNA, their mole percents must be equal. Thus, the mole percent of G, like C, is 20%. The mole percents of A and T account for the remaining 60%. Since A and T always pair, each of their mole percents is equal to half this value: 30%.

5-9. If DNA polymerase requires a perfectly paired primer in order to add the next nucleotide, how is it that any mismatched nucleotides "escape" this requirement and become substrates for mismatch repair enzymes?

Clearly, DNA polymerases must be able to extend a mismatched primer occasionally; otherwise no mismatches would be present in the newly synthesized DNA. Most mismatches are removed by the 3ʹ-to-5ʹ proof- reading exonuclease associated with the DNA polymerase. When the exonuclease does not remove the mismatch, the polymerase can none- theless extend the growing chain. In reality, DNA polymerase and the proofreading exonuclease are in competition with each other. In the case of bacteriophage T7 DNA polymerase, numbers are available that illus- trate this competition. Normally, T7 DNA polymerase synthesizes DNA at 300 nucleotides per second, while the exonuclease removes termi- nal nucleotides at 0.2 nucleotides per second, suggesting that 1 in 1500 (0.2/300) correctly added nucleotides are removed by the exonuclease. When an incorrect nucleotide has been incorporated, the rate of removal increases tenfold to 2.3 nucleotides per second and the rate of polym- erization decreases 3 × 104-fold to 0.01 nucleotide per second. Compari- son of these rates for a mismatched primer suggests that about 1 in 200 (0.01/2.3) mismatched primers will be extended by T7 DNA polymerase.

4-13. Look at the two yeast colonies (mynd í svari). Each of these colonies contains about 100,000 cells descended from a single yeast cell, originally somewhere in the middle of the clump. A white colony arises when the Ade2 gene is expressed from its normal chromosomal location. When the Ade2 gene is moved to a location near a telomere, it is packed into heterochromatin and inactivated in most cells, giving rise to colonies that are mostly red. In these largely red colonies, white sectors fan out from the middle of the colony. In both the red and white sectors, the Ade2 gene is still located near telomeres. Explain why white sectors have formed near the rim of the red colony. Based on the patterns observed, what can you conclude about the propagation of the transcriptional state of the Ade2 gene from mother to daughter cells in this experiment?

Colonies are clumps of cells that originate from a single founder cell and grow outward as the cells divide repeatedly. In the red colony of Figure Q4-3, the Ade2 gene has been inactivated by its position next to the telomere. The inactivation is inherited, but at a low frequency the gene is reactivated. This gives rise to white cells whose descendants are also white (producing the white sectors), even though the gene has not moved away from the telomere. This pattern shows that the inactivation of a telomere-proximal gene is passed on to daughter cells in a way that is not completely stable, and that both the off and the on stare are heritable. An epigenetic mechanism is thought to be involved, based on the tendency of a condensed chromatin state to be inherited following DNA replication.

5-7. DNA repair enzymes preferentially repair mismatched bases on the newly synthesized DNA strand, using the old DNA strand as a template. If mismatches were instead repaired without regard for which strand served as template, would mismatch repair reduce replication errors? Would such a mismatch repair system result in fewer mutations, more mutations, or the same number of mutations as there would have been without any repair at all? Explain your answers.

Mismatch repair normally corrects a mistake in the new strand, using information in the old, parental strand. If the old strand were "repaired" using the new strand that contains a replication error as the template, then the error would become a permanent mutation in the genome, with the "correct" information being erased in the process. erefore, if repair enzymes did not distinguish between the two strands, there would be only a 50% chance that any given replication error would be corrected. Overall, such indiscriminate repair would introduce the same num- ber of mutations as would be introduced if mismatch repair did not exist. In the absence of repair, a mismatch would persist until the next repli- cation. When the replication fork passed the mismatch, and the strands were separated, properly paired nucleotides would be inserted opposite each of the nucleotides involved in the mismatch. A normal, nonmutant duplex would be made from the strand containing the original informa- tion; a mutant duplex would be made from the strand that carried the misincorporated nucleotide. us, the original misincorporation event would lead to 50% mutants and 50% nonmutants in the progeny. is outcome is equivalent to that of indiscriminate repair: averaged over all misincorporation events, indiscriminate repair would also yield 50% mutants and 50% nonmutants among the progeny.

4-14. Mobile pieces of DNA—transposable elements (ísl. stökklar)— that insert themselves into chromosomes and accumulate during evolution make up more than 40% of the human genome. Transposable elements of four types: long interspersed nuclear elements (LINEs), short interspersed nuclear elements (SINEs), long terminal repeat (LTR) retrotransposons, and DNA transposons, are inserted more-or-less randomly throughout the human genome. These elements are conspicuously rare at the four homeobox gene clusters, HoxA, HoxB, HoxC, and HoxD, as illustrated for HoxD (mynd í svari), along with an equivalent region of chromosome 22, which lacks a Hox cluster. Each Hox cluster is about 100 kb in length and contains 9 to 11 genes, whose differential expression along the anteroposterior axis of the developing embryo establishes the basic body plan for humans (and for other animals). Why do you suppose that transposable elements are so rare in the Hox clusters?

The Hox gene clusters are packed with complex and extensive regulatory sequences that ensure the proper expression of individual Hox genes at the correct time and place during development. Insertions of transposable elements into the Hox clusters are eliminated by purifying selection, presumably because they disrupt proper regulation of the Hox genes. Comparison of the Hox cluster sequences in mouse, rat, and baboon reveals a high density of conserved noncoding segments, supporting the presence of a high density of regulatory elements.

4-11. In contrast to histone acetylation, which always correlates with gene activation, histone methylation can lead to either transcriptional activation or repression. How do you suppose that the same modfication—methylation—can mediate different biological outcomes?

The biological outcome associated with histone methylation depends on the site that is modi ed. Each site of methylation has di erent surrounding amino acid context, which allows the binding of distinct reader complexes. It is the binding of different downstream effector proteins that gives rise to different biological outcomes.

4-7. A segment of DNA from the interior of a single strand is shown in Figure Q4-1. What is the polarity of this DNA from top to bottom? (sjá mynd í svari)

The segment of DNA in Figure Q4-1 reads, from top to bottom, 5′-ACT-3′. The carbons in the ribose sugar are numbered clockwise around the ring, starting with C1′, the carbon to which the base is attached, and ending with C5′, the carbon that lies outside the ribose ring.

4-2. (S/Ó) The four core histones are relatively small proteins with a very high proportion of positively charged amino acids; the positive charge helps the histones bind tightly to DNA, regardless of its nucleotide sequence.

True. All the core histones are rich in lysine and arginine, which have basic—positively charged—side chains that can neutralize the negatively charged DNA backbone.

5-4. (S/Ó) When bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand.

True. Consider a single template strand, with its 5ʹ end on the left and its 3ʹ end on the right. No matter where the origin is, synthesis to the left on this strand will be continuous (leading strand), and synthesis to the right will be discontinuous (lagging strand). Thus, when replication forks from adjacent origins collide, a rightward-moving (lagging) strand will always meet a leftward-moving (leading) strand.

4-5. (S/Ó) Gene duplication and divergence is thought to have played a critical role in the evolution of increased biological complexity.

True. Duplication of chromosomal segments, which may include one or more genes, allows one of the two genes to diverge over time to acquire different, but related functions. The process of gene duplication and divergence is thought to have played a major role in the evolution of bio- logical complexity.

5-1. (S/Ó) The different cells in your body rarely have genomes with the identical nucleotide sequence.

True. Each time the genome is copied in preparation for cell division, there is a chance that mistakes (mutations) will be introduced. e rate of mutation for humans is estimated to be 1 nucleotide change per 1010 nucleotides each time the DNA is replicated. Since there are 6.4 × 109 nucleotides in each diploid cell, an average of 0.64 random mutations will be introduced into the genome each time it is copied. us, the two daughter cells from a cell division will often di er from one another and from the parent cell that gave rise to them. Even genomes that are copied perfectly, giving rise to identical daughter cells, will often be altered in subsequent replication cycles. e proportion of identical cells depends on the exact mutation rate.

4-4. (S/Ó) In a comparison between the DNAs of related organisms such as humans and mice, identifying the conserved DNA sequences facilitates the search for functionally important regions.

True. Humans and mice diverged from a common ancestor long enough ago for roughly two out of three nucleotides to have been changed by random mutation. The regions that have been conserved are those with important functions, where mutations with deleterious defects were eliminated by natural selection. Other regions have not been conserved because natural selection cannot operate to eliminate changes in non- functional DNA.

5-2. (S/Ó) In E. coli, where the replication fork travels at 500 nucleotide pairs per second, the DNA ahead of the fork— in the absence of topoisomerase—would have to rotate at nearly 3000 revolutions per minute.

True. If the replication fork moves forward at 500 nucleotide pairs per second, the DNA ahead of it must rotate at 48 revolutions per second (500 nucleotides per second/10.5 nucleotides per helical turn) or 2880 revo- lutions per minute. e havoc this would wreak on the chromosome is prevented by a DNA topoisomerase that introduces transient nicks just in front of the replication fork. is action con nes the rotation to a short, single-strand segment of DNA.

4-1. (S/Ó) Human females have 23 different chromosomes, whereas human males have 24.

True. The human karyotype comprises 22 autosomes and the two sex chromosomes, X and Y. Females have 22 autosomes and two X chromosomes for a total of 23 di erent chromosomes. Males also have 22 autosomes, but have an X and a Y chromosome for a total of 24 di erent chromosomes.

5-3. (S/Ó) In a replication bubble, the same parental DNA strand serves as the template strand for leading-strand synthesis in one replication fork and as the template for lagging-strand synthesis in the other fork.

True. The two ends of a single parental strand of DNA will be copied in the same direction. At the fork at one end of a replication bubble, this will correspond to the leading strand; at the fork at the other end, it will correspond to the lagging strand.

5-8. Discuss the following statement: "Primase is a sloppy enzyme that makes many mistakes. Eventually, the RNA primers it makes are replaced with DNA made by a polymerase with higher delity. This is wasteful. It would be more energy efficient if a DNA polymerase made an accurate copy in the first place.

While the process may seem wasteful, it provides an elegant solution to the di culty of proofreading during primer formation. To start a new primer on a piece of single-strand DNA, one nucleotide must be put in place and then linked to a second and then to a third and so on. Even if these rst nucleotides were perfectly matched to the template strand, such short oligonucleotides bind with very low a nity and it would consequently be di cult to distinguish the correct from incorrect bases by proofreading. e task of the primase is to "just get anything down that binds reasonably well and don't worry about accuracy." Later, these sequences are removed and replaced by DNA polymerase, which uses the accurately synthesized DNA of the adjacent Okazaki fragment as its primer. DNA polymerase has the advantage—which primase lacks—of putting the new nucleotides onto the end of an already existing strand. e newly added nucleotide is held rmly in place, and the accuracy of its base-pairing to the next nucleotide on the template strand can be accu- rately assessed. erefore, as DNA polymerase lls the gap, it can proof- read from the start of the new DNA strand that it makes. What appears at rst glance as energetically wasteful is really just a necessary price to be paid for accuracy.