Chapter 5 - DNA and Chromosomes

Question 1

During the Hershey and Chase experiment, where was the viral protein found?

Answer 1

In the supernatant

Question 2

During the Hershey and Chase experiment, where was the viral DNA found?

Answer 2

In the bacterial pellets

Question 3

Which of the statements below is supported by the Avery, McCarty, and MacLeod experiment and the data in the figure above?

• Lipids from one strain of bacteria can alter the phenotype of another strain if those lipids are taken up by the other strain.
• RNA from one strain of bacteria can alter the phenotype of another strain if that RNA is taken up by the other strain.
• DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain.
• Carbohydrates from one strain of bacteria can alter the phenotype of another strain if those carbohydrates are taken up by the other strain.
• Protein from one strain of bacteria can alter the phenotype of another strain if that protein is taken up by the other strain.

Answer 3

DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain.

(This experiment demonstrated that DNA is the genetic material and is behind the observations previously made by Fred Griffith.)

Question 4

What is the iteration of the Griffith experiment that conclusively demonstrated that transformation must have taken place?

Answer 4

Griffith showed that heat-killed infectious bacteria can transform harmless live bacteria into pathogens.

Question 5

Which option correctly describes the two strands of DNA in a double helix?

• identical
• antiparallel in orientation
• parallel in orientation
• held together by covalent bonds between bases

Answer 5

Antiparallel in orientation.

(The two strands of DNA are considered to be antiparallel in orientation. This means that one strand runs 5’ to 3’ against the 3’-to-5’ orientation of the other strand.)

Question 6

Which structure is normally on the 5’ end of a DNA strand?

• sulfur group
• hydroxyl group
• nitrogenous base
• phosphate group

Answer 6

Phosphate group

(On the 5’ end of a DNA strand, a phosphate group is attached to the ribose sugar. DNA is polar, so the chemistry at the 5’ end differs from the chemistry at the 3’ end. The hydroxyl is attached to the 3’ carbon.)

Question 7

Which chemical group is at the 3’ end of a DNA strand?

• a carboxyl group
• a phosphate group
• a nitrogenous base
• a hydroxyl group

Answer 7

A hydroxyl group

(A hydroxyl group (–OH) is attached to the 3’ carbon of the pentose sugar and (because of the polarity of DNA) is also found at the free, 3’ end of a DNA strand.)

Question 8

What type of bond connects base pairs?

• hydrogen
• covalent
• ionic
• van der Waals

Answer 8

Hydrogen

(A double-stranded DNA molecule is composed of two polynucleotide chains (DNA strands) held together by hydrogen bonds between the paired bases.)

Question 9

The karyotype below was obtained from a human cell. Based on the chromosome spread, what cell type could have been used?

• egg cell
• female human liver cell
• sperm cell
• male human liver cell

Answer 9

Female human liver cell

Homologous chromosomes are present, so the cell is diploid. The two X sex chromosomes indicate female.

Question 10

Prokaryotes have chromosomes that are circular in structure. Which of the following would such chromosomes lack?

• telomeres
• replication origin
• sugar–phosphate backbone
• complementary base pairs
• DNA double helix

Answer 10

Telomeres

(Telomeres are a structure of linear chromosomes that helps prevent the shortening of chromosomes over time due to the replication machinery being unable to attach to the lagging strand of DNA near the end of the chromosome.)

Question 11

In the light microscope, DNA molecules are most visible in which of the following?

• during interphase
• when they are stripped of histones
• when they are in their most extended form
• in the form of euchromatin
• in a cell that is dividing

Answer 11

In a cell that is dividing

(During cell division, the mitotic chromosomes are so highly condensed that individual chromosomes can be seen in the light microscope. Each of these chromosomes contains a single, very long DNA molecule.)

Question 12

In the living cell, histone proteins pack DNA into a repeating array of DNA–protein particles called what?

• octamers
• beads on a string
• euchromatin
• heterochromatin
• nucleosomes

Answer 12

Nucleosomes

(Histones are responsible for the first and most fundamental level of chromatin packing: the formation of the nucleosome.)

Question 13

What is the term that describes the complex of DNA and proteins that makes up a eukaryotic chromosome?

• centromere
• centriole
• chromatin
• centrosome

Answer 13

Chromatin

DNA is wrapped around histone octamers, creating nucleosomes, which are the fundamental unit of chromatin.

Question 14

What structure in an interphase eukaryotic cell is the site of ribosomal RNA transcription?

• ribosome
• nucleolus
• nuclear lamina
• nucleosome

Answer 14

Nucleolus

(During interphase, the parts of different chromosomes that carry genes encoding ribosomal RNAs come together to form the nucleolus.)

Question 15

Which of the following represents the specialized DNA sequence that attaches to microtubules and allows duplicated eukaryotic chromosomes to be separated during M phase?

• nucleosome
• mitotic spindle
• histone
• centrosome
• telomere
• centromere

Answer 15

Centromere

(During mitosis, DNA condenses, adopting a more compact structure and ultimately forming mitotic chromosomes. Once the chromosomes have condensed, the centromere allows the mitotic spindle to attach to each duplicated chromosome in a way that directs one copy of each chromosome to be segregated to each of the two daughter cells.)

Question 16

What are the specialized DNA sequences that are at the ends of most eukaryotic chromosomes called?

• centromeres
• centrosomes
• nucleosomes
• telomeres
• histones

Answer 16

Telomeres

Telomeres contain repeated nucleotide sequences that are required for the ends of chromosomes to be fully replicated.

Question 17

Which of the following is true for most genes?

• A gene is a segment of DNA that contains the instructions for making a particular protein.
• A gene is a segment of DNA that contains the instructions for making a particular RNA.
• A gene is a unit of heredity that contains instructions that dictate the characteristics of an organism.
• All of the above are true regarding genes.
• None of the above are true regarding genes.

Answer 17

All of the above are true regarding genes.

These are all interrelated by the central dogma of molecular biology.

Question 18

What does each eukaryotic chromosome contain?

• one long ribonucleotide sequence
• one single long gene
• one long double-stranded DNA molecule
• one long sequence of amino acids
• one long DNA strand

Answer 18

One long double-stranded DNA molecule

Each eukaryotic chromosome contains one long double-stranded DNA molecule.

Question 19

The tails of the core histone proteins can be chemically modified by the covalent addition of what type of chemical group?

• acetyl
• methyl
• phosphate
• all of these
• none of these

Answer 19

All of these

(The tails of the core histone proteins can be chemically modified by the covalent addition of a methyl group, an acetyl group, or a phosphate group. Each modification alters the physiology of the histone, with some promoting or maintaining euchromatin and others promoting heterochromatin.)

Question 20

What is the general name given to the most highly condensed form of chromatin?

• 30-nm chromatin fiber
• X chromatin
• euchromatin
• nucleosome
• heterochromatin

Answer 20

Heterochromatin

(“Heterochromatin” is the general name given to the most highly condensed form of chromatin, which can be observed under the light microscope.)

Question 21

Determine whether the following statement is true or false:

When a cell divides, its chromatin structure is completely reset.

Answer 21

False.

(This form of “cell memory” is an important feature in how multicellular organisms are able to maintain specific tissue types as the organism grows and differentiates.)

Question 22

Some applications in biology, such as polymerase chain reaction (PCR), require melting the DNA double helix into single strands of DNA. This can be accomplished by heating the DNA. As DNA is heated, why does the double helix structure denature into single strands of DNA but not into individual nucleotides? In other words, why do the single strands remain intact even though the double helix does not?

• The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds.
• The single strands are wrapped around histones, which protects them from denaturing.
• The double helix is held together with phosphodiester bonds, while the single strands are linked by hydrogen bonds.
• The double helix structure is important for DNA replication, while the single strands carry the information.

Answer 22

The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds.

(Hydrogen bonds are weaker than covalent phosphodiester bonds. Hence heating can separate the strands from each other without breaking up the individual strands.)

Question 23

Another step in PCR requires small single-stranded DNA primers to anneal to a target sequence on denatured DNA. Two primers are used. Ideally these primers will have similar melting temperatures. The melting temperature is defined as the temperature at which 50% of the DNA is in the single-stranded form. Which of the following primers will have the highest melting temperature?

• G G G G A A A T T T C C C C
• A A A A G G C C T T T T
• A A A A G G G C C C T T T T
• G G G A A A T T T C C C

Answer 23

G G G G A A A T T T C C C C

(Since G-C base pairs contain one more hydrogen bond than A-T base pairs, it takes more heat energy to separate G-C base pairs than A-T base pairs. This primer, due to its length and G-C content, has the highest melting temperature of the listed primers.)

Question 24

Using this protocol, what should the researchers have seen in terms of the distribution of radioactivity in the centrifuged sample?

• 32P in the pellet, 35S in the solution
• no radioactivity in the pellet
• no radioactivity in the solution
• an equal amount of 35S and 32P in the pellet and the solution
• 35S in the pellet, 32P in the solution

Answer 24

32P in the pellet, 35S in the solution

(Their hypothesis was that either DNA or protein was injected during infection. Because DNA is the genetic material and the viruses inject their genetic material into the bacterial cells, after infection most of the DNA should have been in the bacterial cells in the pellet. Thus, the 32P that was used to label the DNA should have been in the pellet. The head of the virus is composed mainly of protein, so the 35S should have remained with the empty virus heads in the solution.)

Question 25

In the 1920s, bacteriologist Fred Griffith demonstrated that a heat-killed, infectious pneumococcus produced a substance that could convert a harmless form of the bacterium into a lethal one. Fifteen years later, researchers prepared an extract from the disease-causing S strain of pneumococci and showed that this material could transform the harmless R-strain pneumococci cells into the infectious S-strain form. This change to the bacteria was both permanent and heritable, suggesting that this “transforming principle” represents the elusive genetic material of the cells. The researchers subjected their extract to a variety of tests to determine the chemical identity of the “transforming principle.” In one experiment, they treated the material with enzymes that destroy all proteins. This treatment did not affect the ability of the extract to transform harmless bacteria into an infectious form.

From this result, what could the researchers conclude?

• The transforming principle is not DNA.
• The genetic material is not protein.
• The transforming principle is not genetic material.
• Proteins act as the genetic material.
• DNA acts as the genetic material.

Answer 25

The genetic material is not protein.

(From the results, the researchers could conclude that the genetic material is not protein. The results ruled out protein as the genetic material because experimentally destroying the proteins had no effect on the ability of the extract to transform harmless bacteria into an infectious form.)

Question 26

In the late 1920s, bacteriologist Fred Griffith was studying Streptococcus pneumoniae. This bacterium comes in two forms: one that is highly infectious (called the “S strain” because it forms colonies that appear smooth when grown on a nutrient plate in the lab) and one that is relatively harmless (called the “R strain” because its colonies appear rough). When injected into mice, the S strain is lethal, whereas the R strain causes no ill effect. Griffith confirmed that when the S strain is killed by heating, it is no longer infectious. But he then discovered that if he injected mice with both the heat-killed S strain pneumococci and the live, harmless R strain bacteria, the animals died of pneumonia. Furthermore, their blood was swarming with live, S strain bacteria that, when grown in culture, remained infectious and lethal.

Based on these results, what could Griffith conclude?

• The infectious S strain of bacteria cannot be killed by heating.
• The R strain of bacteria is more deadly than previously thought.
• DNA is conclusively confirmed as the genetic material of cells.
• S. pneumoniae is a poor choice for investigating the molecular basis of heredity, as most bacteria behave unpredictably in the laboratory.
• Some substance in the infectious S strain can change the harmless R strain into the more lethal form.

Answer 26

Some substance in the infectious S strain can change the harmless R strain into the more lethal form.

(Based on his results, Griffith could conclude that some substance in the infectious S strain could change the harmless R strain into the more lethal form. The Griffith experiments were performed in the early 1900s, well before the structure of DNA was solved.)

Question 27

In their 1953 paper on the double-helical structure of DNA, Watson and Crick famously wrote: “It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.” What did they mean?

• When a cell divides, each DNA helix is split between the daughter cells.
• The sugar–phosphate backbone of DNA holds the helix together in a way that allows the genetic information to be copied.
• Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand.
• Sexually reproducing organisms swap their DNA

Answer 27

Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand.

(Because each DNA strand in a double helix contains a sequence of nucleotides that is complementary to the sequence of its partner strand, each strand can serve as a template to direct the synthesis of a new strand identical in sequence to its former partner.)

Question 28

One of the first steps in obtaining a karyotype (such as that shown below of a cancer cell) is treating cells with a drug that stalls cells in mitosis. Why must cells arrest in mitosis for karyotype analysis?

• Only mitotic chromosome DNA is separated into single strands, allowing for staining by dyes.
• Only mitotic cells contain homologous chromosomes, because mitosis happens after DNA replication.
• Only mitotic chromosomes are highly condensed and visible with a light microscope.
• Only in mitosis are homologous chromosomes paired up.

Answer 28

Only mitotic chromosomes are highly condensed and visible with a light microscope.

(During interphase, individual chromosomes are not visible with a light microscope. During mitosis, chromosomes become very tightly packed to assist with proper segregation of the chromosomes into the two daughter cells.)

Question 29

Which statement is true about the association of histone proteins and DNA?

• Each histone protein has a deep groove into which a DNA double helix tightly fits.
• Histone proteins insert themselves into the major groove of DNA.
• Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone.
• Histone proteins form hydrogen bonds with the nucleotide bases of DNA.
• Histone proteins have a high proportion of negatively charged amino acids, which bind tightly to the positively charged DNA backbone.

Answer 29

Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone.

(Peripheral arginine and lysine side chains, which are positively charged, interact with the phosphate groups of DNA.)

Question 30

Which of the following statements about nucleosomes is false?

• Nucleosomes convert a DNA molecule into a chromatin thread about one-third the length of the initial DNA.
• Nucleosomes are found only in mitotic chromosomes.
• Nucleosomes can be seen in the electron microscope.
• A nucleosome consists of DNA wrapped around eight histone proteins, plus a short segment of linker DNA.
• Nucleosomes represent the first and most fundamental level of chromatin packing.

Answer 30

Nucleosomes are found only in mitotic chromosomes.

Generally speaking, nucleosomes are found within chromatin regardless of the cell cycle stage.

Question 31

Identify which statement(s) is/are true regarding telomere characteristics.

• Telomeres allow duplicated chromosomes to become separated into daughter cells during M phase.
• Telomeres are found in all living cells.
• Telomeres cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair.
• Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes.
• All of the above are true regarding telomere characteristics.
• Only statements B and D are true regarding telomere characteristics.
• Only statements C and D are true regarding telomere characteristics.

Answer 31

Only statements C and D are true regarding telomere characteristics.

(Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes. These telomeric repeated nucleotide sequences cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair.)

Question 32

What evidence suggests that the large amount of excess “junk” DNA in a genome may serve an important function?

• The nucleotide sequence of “junk” DNA is very similar to that of genes involved in important cell functions.
• Deletions or mutations that fall within “junk” DNA tend to always be harmful to the organism.
• A portion of “junk” DNA is highly conserved in its DNA sequence among many different eukaryotic species.
• The sheer quantity of “junk” DNA suggests it must have some function in the life of the organism.
• All organisms have excess “junk” DNA.

Answer 32

A portion of “junk” DNA is highly conserved in its DNA sequence among many different eukaryotic species.

(Genetic conservation is a significant indicator that there is an important function of a particular sequence.)

Question 33

The Encyclopedia of DNA Elements (ENCODE) project aims to discover functional elements in DNA. One technique employed is DNase-seq. This technique employs a DNase enzyme that digests accessible regions of chromatin, while inaccessible regions remain undigested. The undigested DNA is then sequenced. Which of the following is likely true of the DNase-seq results?

Choose one or more:

• The sequenced DNA will be the same from cell type to cell type.
• Centromeric DNA is likely to be sequenced in all samples.
• Unsequenced DNA regions were likely bound to nucleosomes.
• Unsequenced DNA is likely part of euchromatin.

Answer 33

Centromeric DNA is likely to be sequenced in all samples.
AND
Unsequenced DNA is likely part of euchromatin.

(DNA bound by nucleosomes or tightly packed into heterochromatin will not be digested. The digested sequences are linker DNA between nucleosomes and reflect euchromatin regions of DNA that may be actively transcribed. Since different cell types transcribe different genes, the regions of the genome that are euchromatin vs. heterochromatin may differ between cell types.)

Question 34

How do chromatin-remodeling complexes work?

• They use the energy from ATP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins.
• They use the energy from GTP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins.
• They remove acetyl groups from the tails of histones, rendering the DNA more accessible to other proteins.
• They add methyl groups to the tails of histones in order to attract other proteins.
• They bind to nucleosomes in the 30-nm fiber and induce another level of packing, obscuring DNA from binding by other proteins.

Answer 34

They use the energy from ATP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins.

(Changing the conformation of chromatin is an energy-intensive process and can result in making a segment of DNA more or less accessible to proteins involved, for example, in transcription.)

Question 35

Once heterochromatin has been established, it will often spread until it encounters which of the following?

• a nucleosome
• histone H3 methylated on lysine 9
• a barrier DNA sequence
• euchromatin
• a gene

Answer 35

A barrier DNA sequence

(Barrier sequences harbor their own specific histone modifications that prevent the further propagation of heterochromatin.)

Question 36

Where is heterochromatin not commonly located?

• telomeres
• centromeres
• silenced X chromosomes
• gene-poor regions of chromosomes
• chromosomal regions carrying genes that encode ribosomal proteins

Answer 36

chromosomal regions carrying genes that encode ribosomal proteins

(Chromosomal regions carrying genes that encode ribosomal proteins are active in gene transcription; therefore, you would not expect to see heterochromatin in these areas.)

Question 37

Which of the following statements is not true?

• When a cell divides, its chromatin structures will typically be inherited by its daughter cells.
• A cell will temporarily decondense its chromatin to silence genes during differentiation.
• A cell will temporarily decondense its chromatin to give proteins rapid, localized access to specific DNA sequences.
• A cell will temporarily decondense its chromatin to allow access to specific DNA sequences for replication, repair, or gene expression.
• A cell can permanently condense and silence an entire chromosome during development.

Answer 37

A cell will temporarily decondense its chromatin to silence genes during differentiation.

(Cells will decondense and condense chromatin in patterns that align with gene expression programs, but decondensing chromatin is not a mechanism to silence genes. Rather, the opposite is true: condensing chromatin is a mechanism to silence genes.)