Sapling Ch. 12 Homework

Order the events in which telomerase maintains chromosomal ends during replication. Not all events will be placed.

First -> last The RNA component of telomerase binds to a complementary sequence on the 3′ G overhang of DNA Nucleotides that are complementary to the RNA component of telomerase are added to the 3′ end of the DNA Telomerase moves along the newly synthesized DNA strand toward the 3′ end Additional nucleotides are added to the 3′ end of the DNA Telomerase is removed from the DNA strand entirely Synthesis occurs on the 5′ end of the complementary strand of DNA to fill the gap created by telomerase

Arrange the steps of DNA replication in the order that they occur.

First -> last Helicase unwinds the DNA double helix Single-stranded DNA-binding proteins bind to each template strand RNA primers are added DNA polymerase synthesizes DNA RNA primers are removed DNA ligase joins DNA fragments together

A problem known as the end-replication problem (telomerase problem) exists in eukaryotic chromosomes wherein the chromosomes shorten with each round of DNA replication. Select the statements that best explain why the end-replication problem exists in eukaryotic chromosomes.

DNA polymerase requires a primer for DNA synthesis DNA polymerase synthesizes DNA from the phosphate end to the hydroxyl end

Place the steps of eukaryotic DNA replication in order, from when a germ cell enters gap 1 (G1) phase to the cell cycle termination

Prereplication complex forms at one of many origins of replication The initiation complex creates an active replication fork as helicase unwinds DNA RNA primers are added to provide a 3′ end for elongation 5′ to 3′ synthesis of the leading and lagging strands is carried out by DNA polymerase RNA is replaced with DNA and lagging strands are joined Active telomerase can extend the lost telomere region

Which result did Meselson and Stahl observe by sedimentation equilibrium centrifugation to provide strong evidence for the semiconservative model of DNA replication?

The first generation has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed.

Place the steps of prokaryotic DNA replication in order, from the bacterium initiating DNA replication to DNA-replication termination.

The DNAa protein binds to the single origin of replication The upstream region melts, and helicase binds and unwinds DNA RNA primers are added to provide a 3′ end for elongation DNA polymerase synthesizes the leading strand and short lagging strands 5′ to 3′ DNA replaces RNA and lagging strands are joined Tus protein stops replication forks from crossing each other and ends synthesis

DNA topoisomerase

relaxes the supercoiled DNA

Helicase

unwinds the DNA double helix

SSB protein

prevents the reannealing of DNA strands

DNA polymerase I

replaces RNA primers with DNA

Primase

synthesizes RNA primers

You are studying several alleles of an E. coli helicase gene. One allele, called rsR, confers resistance to RS2014, an antibiotic that works by inhibiting helicase activity. Bacteria with the rsS allele are sensitive to RS2014. Another allele, called ts-, produces a temperature sensitive mutation of the helicase. In bacteria with the ts- allele, helicase is inactivated at 42 degrees C but functions normally at 37 degrees C. Helicase functions normally at both temperatures in bacteria with the ts+ allele.
You obtain the following data while testing four strains of E. coli at different temperatures and doses of RS2014. Each number represents the percentage of maximal DNA synthesis. Based on this data, assign the appropriate genotype to strains A-D in the spaces provided.

Strain A = rsS, ts- Strain B = rsR, ts+ Strain C = rsR, ts- Strain D = rsS, ts+