Chapter 16 - The Molecular Basis of Inheritance

Question 1

Hershey and Chase devised an experiment using radioactive isotopes to determine whether the phage’s DNA or protein entered the bacteria and was the genetic material of T2 phage.

a. How did they label phage protein?
b. How did they label phage DNA?

Separate samples of E. coli were infected with the differently labeled T2 cells, then blended and centrifuged to isolate the bacterial cells from the lighter viral particles.

c. Where was the radioactivity found in the samples with labeled phage protein?
d. Where was the radioactivity found in the samples with labeled phage DNA?
e. What did Hershey and Chase conclude from these results?

Answer 1

a. They grew T2 with E. coli with radioactive sulfur to tag phage proteins.
b. T2 was grown with E. coli in the presence of radioactive phosphorus to tag phage DNA.
c. Radioactivity was found in the supernatant, indicating that the phage protein did not enter the bacterial cells.
d. In the samples with the labeled DNA, most of the radioactivity was found in the bacterial cell pellet.
e. They concluded that viral DNA is injected into the bacterial cells and serves as the hereditary material for viruses.

Question 2

Review the structure of DNA by labeling the following diagram.

Answer 2

a. sugar-phosphate backbone
b. 3’ end of chain
c. hydrogen bonds
d. cytosine
e. guanine
f. adenine (purine)
g. thymine (pyrimidine)
h. 5’ end of chain
i. nucleotide
j. deoxyribose
k. phosphate group

Question 3

Review the structure of DNA by labeling the following diagram.

Answer 3

l. 3.4 nm
m. 0.34 nm
n. 2 nm

Question 4

Using different colors for heavy (parental) and light (new) strands of DNA, sketch the DNA molecules formed in two replication cycles after E. coli were moved from medium containing ¹⁵N to medium containing ¹⁴N. Show the resulting density bands in the centrifuge tubes.

Answer 4

Question 5

Look at the diagram and label the 5’ and 3’ ends of both strands of the DNA molecule.

Answer 5

The phosphate end of each strand is the 5’ end, and the hydroxyl group extending from the 3’ carbon of the sugar marks the 3’ end.

Question 6

In this diagram showing bacterial DNA replication, label the following items: leading and lagging strands, Okazaki fragment, DNA pol III, DNA pol I, DNA ligase, helicase, single-strand binding proteins, primase, RNA primer, and 5’ and 3’ ends of parental DNA.

Answer 6

a. single-strand binding protein
b. DNA pol III
c. leading strand
d. 5’ end of parental strand
e. 3’ end
f. helicase
g. RNA primer
h. primase
i. DNA pol III
j. Okazaki fragment
k. DNA pol I (replacing primer)
l. lagging strand
m. DNA ligase

Question 7

Draw the last Okazaki fragment being formed on the lagging strand of a linear DNA molecule. Indicate how this results in a shortening of the end of the DNA molecule.

Answer 7

No DNA nucleotides can be added after the RNA primer is removed because there is no 3’ end available for DNA polymerase. Thus, the daughter strand is shorter. Further rounds of replication will continue to produce sorter DNA molecules.

Question 8

List the multiple levels of packing in a metaphase chromosome in order of increasing complexity.

Answer 8

nucleosomes (10-nm fiber of ucleosomes and linker DNA)

30-nm fiber

looped domains (300-nm fiber)

coiling and folding of looped domains into highly condensed metaphase chromosome

Question 9

Summarize the evidence and techniques Watson and Crick used to deduce the double-helix structure of DNA.

Answer 9

Watson and Crick used the X-ray diffraction photo of Franklin to deduce that DNA was a helix 2 nm wide, with nitrogenous bases stacked 0.34 nm apart, and making a full turn every 3.4 nm.

Franklin had concluded that the sugar-phosphate backbones were on the outside of the helix with the bases extending inside.

Using molecular models of wire, Watson and Crick experimented with various arrangements and finally paired a purine base with a pyrimidine base, which produced the proper diameter.

Specificity of base pairing (A with T and C with G) is assured by hydrogen bonds.

Question 10

Review your understanding of DNA replication by describing the key enzymes and proteins (in the order of their functioning) that direct replication.

Answer 10

Replication bubbles form where proteins recognize specific base sequences and open up the two strands.

• Helicase*, an enzyme that works at the replication fork, unwinds the helix and separates the strands.
• Single-strand binding proteins* support the separated strands while replication takes place.
• Topoisomerase* eases the twisting ahead of the replication fork by breaking, untwisting, and rejoining DNA strands.
• Primase* synthesizes a primer of about 10 RNA bases to start the new strand.

After a poper base pairs up on the exposed template, DNA polymerase III joins the nucleotide to the 3’ end of the new strand.

On the lagging strand, short Okazaki fragments are formed by primase and polymerase (again moving 5’ → 3’).

• DNA polymerase I* replaces the primer with DNA, adding to the 3’ end of the leading strand or fragments.
• Ligase* joins the 3’ end of one fragment to the 5’ end of its neighbor.

Proofreading enzymes check for mispaired bases, and nucleases, DNA polymerase, and other enzymes repair damage or mismatches.

Question 11

One of the reasons mose scientists believed proteins were the carriers of genetic information was that

a. proteins were more heat stable than nucleic acids.
b. the protein content of duplicating cells always doubled prior to division.
c. proteins were much more complex and heterogeneous molecules than nucleic acids.
d. early experimental evidence pointed to proteins as the hereditary material.
e. proteins were found in DNA.

Answer 11

c. proteins were much more complex and heterogeneous molecules than nucleic acids.

Question 12

Transformation involves

a. the uptake of external genetic material, often from one bacterial strain to another.
b. the creation of a strand of RNA from a DNA molecule.
c. the infection of bacterial cells by phage.
d. the type of semiconservative replication shown by DNA.
e. the replication of DNA along the lagging strand.

Answer 12

a. the uptake of external genetic material, often from one bacterial strain to another.

Question 13

The DNA of an organism has thymine as 20% of its bases. What percentage of its bases would be guanine?

a. 20%
b. 30%
c. 40%
d. 60%
e. 80%

Answer 13

b. 30%

Question 14

In his work with pneumonia-casing bacteria, Griffith found that

a. DNA was the transforming agent.
b. the pathogenic and harmless strains mated.
c. heat-killed harmless cells could cause pneumonia when mixed with heat-killed pathogenic cells.
d. some heat-stable chemical was transferred to harmless cells to transform them into pathogenic cells.
e. a T2 phage transformed harmless cells to pathogenic cells.

Answer 14

d. some heat-stable chemical was transferred to harmless cells to transform them into pathogenic cells.

Question 15

T2 phage is grown in E. coli with radioactive phosphorus and then allowed to infect other E. coli. The culture is blended to separate the viral coats from the bacterial cells and centrifuged. Which of the following best describes the expected results of such an experiment?

a. Both viral and bacterial DNA are labeled; radioactivity is found in the supernatant.
b. Viral DNA is labeled; radioactivity is found in the pellet.
c. Viral proteins are labeled; radioactivity is found in the supernatant but not in the pellet.
d. Both viral and bacterial proteins are labeled; radioactivity is present in both the supernatant and the pellet.
e. The virus destroyed the bacteria; no pellet is formed.

Answer 15

b. Viral DNA is labeled; radioactivity is found in the pellet.

Question 16

In their classic experiment, Meselson and Stahl

a. provided evidence for the semiconservative model of DNA replication.
b. were able to separate phage protein coats from E. coli by using a blender.
c. found that DNA labeled with ¹⁵N was of intermdiate density.
d. grew E. coli on labeled phosphorus and sulfur.
e. found that DNA composition was species specific.

Answer 16

a. provided evidence for the semiconservative model of DNA replication.

Question 17

Watson and Crick concluded that each base could not pair with itself because

a. there would not be room for the helix to make a full turn every 3.4 nm.
b. the uniform width of 2 nm would not permit two purines or two pyrimidines to pair together.
c. the bases could not be stacked 0.34 nm apart.
d. identical bases could not hydrogen-bond together.
e. they would be on antiparallel strands.

Answer 17

b. the uniform width of 2 nm would not permit two purines or two pyrimidines to pair together.

Question 18

The joining of nucleotides in the polymerization of DNA requires energy from

a. DNA polymerase.
b. the hyrolysis of the terminal phosphate group of ATP.
c. RNA nucleotides.
d. the phosphate groups of the sugar-phosphate backbone.
e. the hydrolysis of the pyrophosphates removed from nucleoside triphosphates.

Answer 18

e. the hydrolysis of the pyrophosphates removed from nucleoside triphosphates.

Question 19

The continuous elongation of a new DNA strand along one strand of DNA

a. requires the action of DNA ligase as well as polymerase.
b. occurs because DNA ligase can only elongate in the 5’ → 3’ direction.
c. occurs on the leading strand.
d. occurs on the lagging strand.
e. a, b, and c are correct.

Answer 19

c. occurs on the leading strand.

Question 20

Which of the following statements about DNA polymerase is incorrect?

a. It forms the bonds between complementary base pairs.
b. It is able to proofread and correct for errors in base pairing.
c. It is unable to initiate synthesis; it requires an RNA primer.
d. It only works in the 5’ → 3’ direction.
e. It is found in eukaryotes and prokaryotes.

Answer 20

a. It forms the bonds between complementary base pairs.

Question 21

Thymine dimers — covalent links between adjacent thymine bases in DNA — may be induced by UV light. When they occur, they are repaired by

a. excision enzymes (nucleases).
b. DNA polymerase.
c. ligase.
d. primase.
e. a, b, and c are all needed.

Answer 21

e. a, b, and c are all needed.

Question 22

How does DNA synthesis along the lagging strand differ from that on the leading strand?

a. Nucleotides are added to the 5’ end instead of the 3’ end.
b. Ligase is the enzyme that polymerizes DNA on the lagging strand.
c. An RNA primer is needed on the lagging strand but not on the leading strand.
d. Okazaki fragments, which each grow 5’ → 3’, must be joined along the lagging strand.
e. Helicase synthesis Okazaki fragments, which are then joined by ligase.

Answer 22

d. Okazaki fragments, which each grow 5’ → 3’, must be joined along the lagging strand.

Question 23

Which of the following enzymes or proteins is paired with an incorrect or inaccurate function?

a. Helicase — unwind and separate parental double helix
b. Telomerase — add telomere repetitions to ends of chromosomes
c. Single-strand binding protein — hold strands of unwound DNA apart and straight
d. Nuclease — cut out (excise) damaged DNA strand
e. Primase — form DNA primer to start replication

Answer 23

e. Primase — form DNA primer to start replication

Question 24

Use the diagram to answer the question:

Which letter indicates the 5’ end of this single DNA strand?

a. b. c. d. e.

Answer 24

a.

Question 25

Use the diagram to answer the question:

At which letter would the next nucleotide be added?

a. b. c. d. e.

Answer 25

d.

Question 26

Use the diagram to answer the question:

Which letter indicates a phosphodiester bond formed by DNA polymerase?

a. b. c. d. e.

Answer 26

b.

Question 27

Use the diagram to answer the question:

The base sequence of the DNA strand made from this template would be (from top to bottom)

a. A T C.
b. C G A.
c. T A C.
d. U A C.
e. A T G.

Answer 27

c. T A C.

Question 28

What are telomeres, and what do they do?

a. ever-shortening tips of chromosomes that may signal cells to stop dividing at maturity
b. highly repetitive sequences at tips of chromosomes that protect the lagging strand during replication
c. repetitive sequences of nucleotides at the centromere region of a chromosome
d. enzymes that are present in germ cells that allow these cells to undergo repeated divisions
e. Both a and b are correct.

Answer 28

e. Both a and b are correct.

Question 29

You are trying to support your hypothesis that DNA replication is conservative (i.e., parental strands separate; complementary strands are made, but these new strands join together to make a new DNA molecule, and the parental strands rejoin.)

You take E. coli that had grown in a medium containing only heavy nitrogen (¹⁵N) and transfer a sample to a medium containing light nitrogen (¹⁴N).

After allowing time for only one DNA replication, you centrifuge a sample and compare the density band(s) formed with control bands for bacteria grown on either normal ¹⁴N or ¹⁵N medium.

Which band location would support your hypothesis of conservative DNA replication?

Answer 29

c.

Question 30

You are trying to support your hypothesis that DNA replication is conservative (i.e., parental strands separate; complementary strands are made, but these new strands join together to make a new DNA molecule, and the parental strands rejoin.)

You take E. coli that had grown in a medium containing only heavy nitrogen (¹⁵N) and transfer a sample to a medium containing light nitrogen (¹⁴N).

After allowing time for only one DNA replication, you centrifuge a sample and compare the density band(s) formed with control bands for bacteria grown on either normal ¹⁴N or ¹⁵N medium.

Which centrifuge tube would represent the band distribution obtained after one replication showing the DNA replication is semiconservative?

Answer 30

a.

Question 31

Place the following structures in order from smallest to largest. Then choose the structure that is in the middle of that size range.

a. looped domain
b. histone H1
c. nucleosome
d. 30-nm fiber
e. metaphase chromosome

Answer 31

d. 30-nm fiber

Question 32

Which of the following would not be involved in the proper behavior of chromosomes in meiosis?

a. detachment of looped domains from the nuclear lamina
b. phosphorylation of particular amino acids in histone tails
c. condensing of centromeres and telomeres, converting them from euchromatin to heterochromatin
d. coiling and folding of looped domains
e. All of the above are involved in the behavior of chromosomes in meiosis.

Answer 32

c. condensing of centromeres and telomeres, converting them from euchromatin to heterochromatin