mastering ch. 11

Retrotransposons

Insert into new locations by first being transcribed to create RNA copies, which are then reverse-transcribed into DNA.

how much DNA is present in the photo? there appears to be a span of 13 "beads" present.

Since there are 13 nucleosome core particles connected by 12 linker regions: (13 x 146) + (12 x 54bp of liker dna) = 2546 bp

497 sepia, fully fertile 493 wild-type, semi-sterile 4 sepia, semi-sterile 6 wild-type, fully fertile

Since these two recombinant products have a frequency of 1% in the F2 (calculated as follows: (4 + 6)/1000 = 0.01), you can infer that there is a distance of 1 map unit between the sepia locus and the translocation breakpoint on chromosome III. 1 map unit

F1 males x pure black females (Note: There is no crossing over in the F1 males.) F1 females x pure black males (Note: Crossing over may occur in the F1 females.) F2 Progeny 1/2 black, fully fertile 1/2 wild-type, semi-sterile 424 black, fully fertile 426 wild-type, semi-sterile 73 black, semi-sterile 77 wild-type, fully fertile

Since these two recombinant products have a frequency of 15% in the F2 (calculated as follows: (73 + 77)/1000 = 0.15), you can infer that there is a distance of 15 map units between the black locus and the translocation breakpoint on chromosome II. 15 map units

Mismatch repair corrects DNA replication mistakes by removing the incorrect nucleotide of a mismatched ______ and then ________ that part of the DNA molecule. The incorrect nucleotide of a mismatched________ is on ______________________ strand and the correct nucleotide is on _____________________. . If the identification of new versus template strands was not possible, then the nucleotide removed from the mismatched________ would be random, causing mutation instead of repair half of the time. The template strand is identified in bacteria_______________________

base pair re-replicating base pair the newly synthesized, daughter DNA the parental, template DNA strand base pair as the strand that contains methylated nucleotides

It has been observed that diseases caused by repeat expansions are the result of trinucleotide repeats rather than smaller or larger repeat lengths. If non-trinucleotide repeat expansions were present in the coding region of a gene, which type of mutation would you expect to be the most likely?

frameshift Repeat lengths that are not multiples of three in the coding regions of genes would likely cause frameshift mutations, rendering the protein product nonfunctional. Two mutant alleles may not be developmentally viable; thus these genotypes do not survive to pass on these alleles.

translocation

moving of a segment to a to another chromosome. reciprocal- 2 chromosomes exchange segments or non reciprocal- just one transfer

Identify two DNA repair mechanisms that remove UV-induced DNA lesions.

nucleotide excision photoreactivation

duplication

repeats a segment it can be right next to the original location or in a new location orientation may be the same or the reverse

Inversion

the removal of a segment and then reinserted in the same chromosome but in reverse orientation

Which two repair processes are the most error prone?

translesion DNA synthesis nonhomologous end joining Both double-strand breaks and DNA lesions can block replicative DNA polymerases from duplicating the genome which can stall the cell cycle. When other repair mechanisms are unable to restore the genome to an intact state, repair mechanisms that allow for DNA sequence changes can at least restore the genome to a state that replicative DNA polymerase can proceed. These processes are said to be error prone. Nonhomologous end joining repairs double-strand breaks, but in doing so nucleases have to trim back the DNA surrounding the break. Thus DNA sequences are lost, but the break is ligated back together. Translesion DNA synthesis is performed by bypass polymerases. Lesions that block replicative DNA polymerases do not block bypass polymerases. In order to synthesize DNA through lesions, these polymerases lack proofreading activity. Thus, DNA sequence errors can be incorporated into the DNA.

What kinds of gene mutations are induced by radiation energy?

transversions transitions

Assume that the DNA associated with a nucleosome core particle plus the DNA in the linker adds up to 200 bp. Approximately how many base pairs are found in the linker region?

54 Approximately 146 bp of DNA is wrapped around the nucleosome core particle. Therefore 200 bp - 146 bp = 54 bp of linker DNA. This is the amount of DNA present in the "string."

If a DNA replication error is detected by DNA polymerase, how is it corrected?

DNA polymerase removes the last nucleotide added using its 3'-to-5' exonuclease activity and adds the correct nucleotide in its place.

Transposons

Insert into new locations by excising themselves from one location and inserting into a new location. Insert into new locations by DNA replication and insertion of the replicated copy into a new location

The process of DNA double-strand breaks and repair in bacteria and during homologous recombination in eukaryotes are _______ in many respects. In both processes, a strand of DNA from the _______ DNA molecule invades the DNA ________of an intact, homologous sequence in a process called ________

similar broken duplex strand invasion

How many total polypeptide chains would be present in a single nucleosome core particle?

8 Two molecules each of four histones - H2A, H2B, H3, and H4 - join together to form 8 nucleosome core particle.

Which histone helps stabilize the solenoid structure?

H1 (it is a linker) plays a key role in stabilizing the 30-nm solenoid structure. The long N-terminal and C-terminal ends of the H1 protein attach to adjacent nucleosome core particles pulling them into an orderly solenoid array.

__________is a light-dependent mechanism involving photolyase, an enzyme present in many microbes that ______the cross-links created by UV irradiation and_____ the normal DNA structure. _________ repairs involves _______of the DNA damage site, _______ of a segment of the damaged strand, and after that DNA _________using the complimentary stand to fill in the gap and ______ to to seal the nick between the newly synthesized DNA and the original DNA strand.

Photoreractivation removes restores Nucleotide excision recognition excision synthesis ligation

Strand invastion results in a structure termed a D-loop___________________and DNA synthesis occurs using the _________ homologous DNA strands as template. The resulting DNA structure resembles __________ and is resolved to create a repaired DNA molecule that contains the DNA sequence of the homolog at the site of repair.

in both bacteria and eukaryotes intact a Holliday junction

What kind of DNA lesion does UV energy cause?

pyrimidine dimerisation

If thymidine dimers created by UV irradiation persist until DNA replication, DNA polymerase is likely to make a mistake reading ____________ and insert _________ on a the complementary strand. The altered sequence on the newly synthesized strand _________ during the next round, or replication, thus creating __________

the template sequence at TT something besides AA will be copied a mutant sequence

nondisjunction (2n=6) ... -meiosis normally -in 1 chromo pair in meiosis 1 -in all 3 chromo pairs in meosis 1 (all go to 1 pole) -of 1 chromo in 1 daughter cell in meiosis 2 -in all 3 chromo in 1 daughter cell in meiosis 2

3 2 or 4 0 or 6 2, 3, or 4 0, 3, or 6

Splice site mutation Frameshift mutation

changes several amino acids in a protein and results in a protein that is shorter than the wild-type product

ince sea urchins have longer linker DNA segments than Drosophila melanogaster, when this region of DNA is moved to the sea urchin it is likely that these enhancer sequences will ______________________ With enhancer sequences available in sea urchins to bind regulatory proteins that stimulate transcription, the same region of DNA will likely generateRNA transcripts.

fall within linker DNA. more more

If a replication error escapes detection and correction, what kind of abnormality is most likely to exist at the site of replication error? Identify two mechanisms that can correct the kind of abnormality If the kind of abnormality is not corrected before the next DNA replication cycle, what kind of mutation occurs

mismatched base pairs proofreading by DNA polymerase mismatch repair transition mutation

Which repair process in E. coli uses visible light to repair thymine dimers?

photoreactivation repair Photoreactivation utilizes the enzyme photolyase to break the bonds formed during pyrimidine dimerization. A thymine dimer produced by UV irradiation is bound by photolyase. Visible light energy is absorbed by photolyase and is redirected to break the bonds forming the dimer. In addition to bacteria, this DNA repair mechanism is also found in single-celled eukaryotes, plants, and some animals such as Drosophila.

What general mechanism do DNA polymerases use to check the accuracy of DNA replication and identify errors during replication?

They remove incorrect nucleotides immediately after they are added. They ensure that the correct nucleotide is added to the growing DNA strand.

The length of linker DNA segments varies somewhat among organisms. The length of the linker DNA in sea urchin is 110 base pairs. Suppose it were possible to move the entire region of DNA (92,672 base pairs long) from Drosophila melanogaster into the genome of a sea urchin. Approximately how many nucleosomes would be required to organize this region into the 10nm fiber structure?

146+110=362 /92672= 362

What are the examples of chemical mutagens?

EMS hydroxylamine

Histones in general have a net positive charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and weakens the histone-DNA interaction.

Histones are basic proteins that interact with negatively charged DNA. The strength of this interaction is modulated by epigenetic modifications. One of these modifications is acetylation. Acetylation adds an acetyl group to the positively charged amino group present on the side chain of the amino acid lysine effectively changing the net charge of the protein by neutralizing the positive charge. When the positive charge is reduced, the histones loosen their grip on the negatively charged DNA.

What are the examples of radiant energy that cause mutations?

UVradiation γ radiation

deletion

loss of part of a chromosomal segment

DNA mismatch repair can accurately distinguish between the template strand and the newly replicated strand of a DNA duplex. What characteristic of DNA strands is used to make this distinction?

methylation

How are his− bacteria used in the Ames test? What mutational event is identified using his− bacteria? The test detects _________using his− bacteria, which are bacteria that _________on medium that lacks the amino acid histidine. Reversion of his− back to his+ is detected by plating millions of his− bacteria on medium_________ and counting the number of colonies that grow.

reverse mutations cannot grow lacking histidine

The second affects one of several duplicate tRNA genes. This base-pair substitution mutation changes the anticodon sequence of a tRNATrp from 3'-ACC - 5' to 3'-ACU - 5'. Do you consider the second mutation to be a forward mutation or a reversion?

second-site reversion (suppressor)

Describe the purpose of the Ames test.

to determine whether a chemical can act as a mutagen

Approximately how many nucleosomes are required to organize this region of DNA into the 10nm fiber structure? Drosophila melanogaster‎ is 92,672 base pairs longThe length of linker DNA in Drosophila‎ is 35 base pairs.

146bp +35=181 /92672=512 nucleosomes

Rank the following levels of chromatin compaction in eukaryotes from the least compact to the most compact. solenoid, chromaid, metaphase chromosome, loop domains, naked DNA, nuclesome

Chromatin compaction is required for the nucleus to accommodate genomes that are often more than 1,000 times longer than the nuclear diameter. (2nm DNA, 11nm nucleosome, 30nm solenoid, 300nm loop domains, 700nm chromatid, and 1400nm metaphase chromosome).

Chromatin structure is dynamic. In regions of highly condensed chromatin, such as the centromere, the boundary between heterochromatin and euchromatin is variable. Genes that are near this boundary region can be influenced by either type of chromatin in what is referred to as position effects. Recall the early Drosophila melanogaster experiments by Hermann Muller where the repositioning of the w + allele (normal activity of the w + allele produces red eye pigment) by translocation or inversion near this boundary of chromatin produced intermittent w + activity. In the heterozygous state (w +/w), a variegated eye is produced, with white and red patches.

Even though eye color phenotypic variegation exists within the eye, all cells have the same genotype. When heterochromatin spreading encompasses the new location of w + allele, the gene is not transcribed, producing white eye patches. When heterochromatin spreading does not reach the new location of the w + allele, the gene will be transcribed, producing red eye patches.

If this trisomic fly is the progeny of a male fly of genotype (ey+ ey−, gw+ gw−) crossed to a female fly with genotype (ey− ey−, gw− gw−), what can you conclude about the events that led to its formation? Nondisjunction occurred in the male parent at the meiosis I division to produce an (n+1) sperm that fused with a normal (n) egg. Nondisjunction occurred in the female parent at the meiosis II division Nondisjunction occurred in the male parent at the meiosis II division Nondisjunction occurred in the female parent at the meiosis I division

Nondisjunction occurred in the male parent at the meiosis I division to produce an (n+1) sperm that fused with a normal (n) egg. The (ey−gw−) chromosome in the trisomic male must have come from the female parent, and the other two chromosomes must have come from the male parent. Since these two chromosomes have distinct alleles of ey and gw, they were not sister chromatids (i.e., replicated copies of the same chromosome). That means that they must have been paired as homologous chromosomes at synapsis, and remained in the same cell through a nondisjunction event at the meiosis I division.

A 1-mL sample of the bacterium E. coli is exposed to ultraviolet light. The sample is used to inoculate a 500-mL flask of complete medium that allows growth of all bacterial cells. The 500-mL culture is grown on the benchtop, and two equal-size samples are removed and plated on identical complete-medium growth plates. Plate 1 is immediately wrapped in a dark cloth, but plate 2 is not covered. Both plates are left at room temperature for 36 hours and then examined. Plate 2 is seen to contain many more growing colonies than plate 1. Thinking about DNA repair processes, how do you explain this observation?

The level of DNA damage inflicted by UV light was high enough to require the photoreactive repair pathway to repair the DNA damage sufficiently to prevent cell death.

Missense mutation

produces a mutant protein that differs from the wild-type protein at one amino acid position

Nonsense mutation

produces a protein that is shorter than the wild-type protein but does not have any amino acid changes in the portion produced

Promoter mutation

produces about 5% of the wild-type amount of an mRNA

The partial amino acid sequence of a wild-type protein is: . . . Arg-Met-Tyr-Thr-Leu-Cys-Ser . . . The same portion of the protein from a mutant has the sequence: . . . Arg-Thr-Tyr-Thr-Leu-Cys-Ser . . . What type of mutation?

substitution mutation

Which repair process(es) use(s) a DNA polymerase?

translesion DNA synthesis base excision repair nucleotide excision repair homologous recombination (synthesis-dependent strand annealing)

What kinds of gene mutations are induced by chemical mutagens?

transversions transitions frameshifts

Two different mutations are identified in a haploid strain of yeast: The first prevents the synthesis of adenine by a nonsense mutation of the ade-1 gene. In this mutation, a base-pair substitution changes a tryptophan codon (UGG) to a stop codon (UGA). The second affects one of several duplicate tRNA genes. This base-pair substitution mutation changes the anticodon sequence of a tRNATrp from 3'-ACC - 5' to 3'-ACU - 5'. Assuming there are no other mutations in the genome, will this double-mutant yeast strain be able to grow on minimal medium?

yeah