Chapter 16 Critical Thinking

A mouse receives living R bacteria and survives. What does this show?

The R strain is nonvirulent.

A mouse receives heat-killed S bacteria and survives. What does this show?

Dead S bacteria cannot cause disease by themselves.

A mouse receives heat-killed S bacteria plus living R bacteria and dies. What major conclusion follows?

Hereditary information from S cells transformed R cells.

Why did living S bacteria appear in the dead mouse after Griffith's mixed treatment?

Living R cells acquired DNA that allowed capsule production.

If DNase is added to heat-killed S bacteria before mixing with living R bacteria, what result is expected?

No transformation and no living S bacteria.

If protease is added to heat-killed S bacteria before mixing with living R bacteria, what result is expected?

Transformation can still occur.

Why did Avery's DNase result provide stronger evidence than Griffith's experiment alone?

It identified DNA specifically as the transforming material.

Why was sulfur-35 an effective protein label in the Hershey-Chase experiment?

Protein contains sulfur-containing amino acids, while DNA lacks sulfur.

Why was phosphorus-32 an effective DNA label in the Hershey-Chase experiment?

DNA has phosphorus in its phosphate backbone.

If sulfur-35 is found mainly in the supernatant after blending and centrifugation, what does that mean?

Protein coats remained outside bacterial cells.

If phosphorus-32 is found mainly in the pellet after blending and centrifugation, what does that mean?

DNA entered bacterial cells.

Why did the Hershey-Chase experiment support DNA rather than protein as genetic material?

DNA entered bacteria and was associated with production of new phages.

A DNA sample has 30% adenine. What percentage is thymine?

30%.

A DNA sample has 30% adenine. What percentage is guanine?

20%.

A DNA sample has 18% guanine. What percentage is cytosine?

18%.

A DNA sample has 18% guanine. What percentage is adenine?

32%.

A double-stranded DNA molecule has 40% cytosine. What percentage is adenine?

10%.

Why would an equal amount of adenine and guanine not necessarily be expected in DNA?

Chargaff's rules require A equals T and G equals C, not A equals G.

What is the complementary DNA strand for 5′-ATGCC-3′?

3′-TACGG-5′.

What is the complementary DNA strand written 5′ to 3′ for 5′-ATGCC-3′?

5′-GGCAT-3′.

Why must the complement of 5′-ATGCC-3′ be written 3′-TACGG-5′ before reversing it?

DNA strands are antiparallel.

A DNA molecule is 3′-TACGGA-5′. What new DNA strand can be synthesized from it?

5′-ATGCCT-3′.

A mutation replaces one G-C pair with one A-T pair. How does the hydrogen-bond number change?

It decreases by one hydrogen bond.

Why does G-C-rich DNA usually require more heat to separate than A-T-rich DNA?

G-C pairs have three hydrogen bonds instead of two.

What would the conservative model predict after one generation in nitrogen-14 medium?

One heavy band and one light band.

What would the dispersive model predict after two generations in nitrogen-14 medium?

One band of intermediate density that becomes progressively lighter.

What did semiconservative replication predict after one generation in nitrogen-14 medium?

All DNA would have intermediate density.

What did semiconservative replication predict after two generations in nitrogen-14 medium?

Half intermediate DNA and half light DNA.

Why did the first Meselson-Stahl generation rule out conservative replication?

Conservative replication predicts separate heavy and light bands, not one intermediate band.

Why did the second Meselson-Stahl generation rule out dispersive replication?

Dispersive replication predicts one increasingly light intermediate band, not separate light and hybrid bands.

After three generations in nitrogen-14 medium, what fraction of DNA molecules should remain hybrid under semiconservative replication?

One-fourth.

After three generations in nitrogen-14 medium, what fraction of DNA molecules should be light under semiconservative replication?

Three-fourths.

Why are multiple origins especially important in eukaryotic DNA replication?

Large linear chromosomes would take too long to copy from one origin.

Why does a circular bacterial chromosome not have the same end-replication problem as a linear eukaryotic chromosome?

Circular DNA has no chromosome ends.

A replication fork cannot unwind DNA. Which enzyme is likely defective?

Helicase.

A replication fork has excessive twisting and strain ahead of it. Which enzyme is likely defective?

Topoisomerase.

Separated DNA strands rapidly rejoin during replication. Which proteins are likely missing?

Single-strand binding proteins.

DNA replication begins but cannot start new DNA strands. Which enzyme is likely defective?

Primase.

RNA primers remain embedded in the completed bacterial DNA. Which enzyme is likely defective?

DNA polymerase I.

Okazaki fragments are present but remain disconnected. Which enzyme is likely defective?

DNA ligase.

A cell makes a continuous new strand but no short fragments. Which strand is most directly affected?

The lagging strand.

Why is the leading strand synthesized continuously?

Its template runs 3′ to 5′ toward the replication fork.

Why is the lagging strand synthesized discontinuously?

Its template orientation forces synthesis away from the fork in short fragments.

Which direction does DNA polymerase add nucleotides to every new DNA strand?

5′ to 3′.

Why can DNA polymerase not synthesize a strand in the 3′ to 5′ direction?

It can add nucleotides only to a free 3′ hydroxyl group.

A newly synthesized strand has an incorrect base that escaped proofreading. What repair pathway may correct it later?

Mismatch repair.

A UV-exposed cell has bulky thymine dimers. What repair pathway is most appropriate?

Nucleotide excision repair.

Why can failure of DNA repair increase cancer risk?

Persistent mutations can alter genes that control cell division.

Why does loss of proofreading increase mutation rate?

Incorrect nucleotides remain in newly copied DNA.

Why can a telomerase inhibitor limit cancer-cell division?

Cancer cells may no longer maintain telomere length.

Why can excessive telomerase activity contribute to cancer?

It can allow cells to avoid normal limits on repeated division.

Why might a cell with short telomeres stop dividing?

Further replication risks losing important DNA near chromosome ends.

A region of DNA is tightly condensed and poorly transcribed. Is it likely euchromatin or heterochromatin?

Heterochromatin.

A region of DNA is loosely packed and actively transcribed. Is it likely euchromatin or heterochromatin?

Euchromatin.

Why can two cells with identical DNA express different genes?

Their chromatin packing and gene accessibility can differ.

Why are histones important for eukaryotic DNA?

They package long DNA molecules into a compact, organized chromosome structure.

Why are chromosome territories biologically useful?

They organize chromosomes within the nucleus instead of leaving them as a random tangled mass.