Author: Biology Junction Team

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Multiple Choice
Identify the choice that best completes the statement or answers the question.

Mendel verified true-breeding pea plants for certain traits before undertaking his experiments. The term "true-breeding" refers to:

a.	genetically pure lines.
b.	organisms that have a high rate of reproduction.
c.	organisms that will produce identical copies of themselves upon reproduction.
d.	organisms that are heterozygous for a given trait.
e.	organisms that are homozygous for all possible traits.

Mating a true-breeding pink rose plant with a true-breeding pink rose plant will produce:

a.	plants with pink, red, and white roses.
b.	only plants with pink roses.
c.	plants with red or white roses in a 3:1 ratio.
d.	plants with white or red roses in a 3:1 ratio.
e.	None of the above.

The term "dominant" means that:

a.	both alleles can be expressed in a hybrid.
b.	all members of the F2 generation of a hybrid cross exhibit the dominant phenotype.
c.	one allele can mask the expression of another in a hybrid.
d.	the dominant phenotype shows up in 100% of the offspring in all generations.
e.	the dominant phenotype is more beneficial than the recessive phenotype.

Mendel's principle of segregation states that:

a.	alleles from one parent mask the expression of alleles from the other parent.
b.	alleles separate from each other before forming gametes.
c.	hybrids will express a phenotype intermediate between the two parental phenotypes.
d.	true-breeding parents produce offspring of the same phenotype.
e.	different loci separate from each other.

A pear plant with the genotype Aa can produce gametes containing:

a.	either A or Aa.
b.	only the dominant A.
c.	only the recessive a.
d.	either A or a.
e.	either AA, Aa or aa.

The physical appearance of an organism for a given trait is termed:

a.	genetics.
b.	dominance.
c.	synapsis.
d.	genotype.
e.	phenotype.

The physical location of a particular gene on a chromosome is called:

a.	an allele.
b.	a locus.
c.	a trait.
d.	a chromatid.
e.	None of the above.

____________ are alternative forms of a gene that govern the same feature, such as eye color, and occupy corresponding positions on homologous chromosomes.

a.	Alleles
b.	Loci
c.	Homozygotes
d.	Coupled traits
e.	None of the above.

Using standard conventions for naming alleles, which of the following pairs is correct?

a.	Tt—recessive phenotype
b.	TT—heterozygous
c.	tt—homozygous
d.	tt—dominant phenotype
e.	All of the above.

The separation of alleles of a gene takes place during:

a.	anaphase of mitosis.
b.	cytokinesis of mitosis.
c.	anaphase I of meiosis.
d.	telophase II of meiosis.
e.	cytokinesis of meiosis.

Which of the following represents the possible genotype(s) resulting from a cross between an individual homozygous (BB) and one heterozygous (Bb) individual?

a.	BB and Bb
b.	BB, Bb, and bb
c.	BB only
d.	Bb only
e.	bb only

Which of the following represents the possible genotype(s) resulting from a cross between two individuals that are heterozygous (Bb)?

a.	BB and Bb
b.	BB, Bb, and bb
c.	BB only
d.	Bb only
e.	bb only

Which of the following represents the possible genotype(s) resulting from a cross between an individual heterozygous (Bb) and one that is homozygous (bb)?

a.	BB and Bb
b.	Bb, and bb
c.	BB only
d.	Bb only
e.	bb only

Mating an individual expressing a dominant phenotype, but whose genotype is unknown, with an individual expressing the corresponding recessive phenotype is an example of:

a.	a heterozygous cross.
b.	an F1 cross.
c.	an F2 cross.
d.	a parental cross.
e.	a test cross.

The genotype for a pea plant that is homozygous recessive for both height and pea color would be:

a.	tt.
b.	YY.
c.	TtYy.
d.	ttyy.
e.	TTYY.

In peas, Mendel found that tall plants and yellow peas are dominant. The phenotype for a pea plant with the genotype TTyy would be:

a.	heterozygous.
b.	Ty.
c.	short with yellow peas.
d.	tall with green peas.
e.	tall with yellow peas.

In peas, Mendel found that tall plants and yellow peas are dominant. The phenotype for a pea plant with the genotype TtYy would be:

a.	heterozygous.
b.	Ty.
c.	short with yellow peas.
d.	tall with green peas.
e.	tall with yellow peas.

The height of pea plants from a cross between parent plants heterozygous for height, in which tall is dominant, would be:

a.	all short.
b.	all tall.
c.	1 tall : 3 short.
d.	2 short : 2 tall.
e.	3 tall : 1 short.

Which of the following represents the possible genotype(s) resulting from a cross between an individual homozygous for black hair (BB) and an individual homozygous for blonde hair (bb)?

a.	BB and Bb
b.	BB, Bb, and bb
c.	BB only
d.	Bb only
e.	bb only

What is the probability that two lizards that are heterozygous for stripes on their tails (Ss) will produce an offspring that is homozygous for no stripes (ss)?

a.	1
b.	1/2
c.	1/4
d.	1/8
e.	0

In humans, assume that brown eyes is dominant and blue eyes is recessive. If two brown-eyed individuals have a child with blue eyes, that means:

a.	both parents are homozygous for brown eyes.
b.	both parents are heterozygous for eye color.
c.	there is a 1/4 chance that their second child will have brown eyes.
d.	there is a 50/50 chance that their second child will have blue eyes.
e.	None of the above.

If a couple is planning on having two children, what is the probability that both will be male?

a.	0
b.	1/4
c.	1/2
d.	3/4
e.	1

If a couple is planning on having three children, what is the probability that only one will be male?

a.	0
b.	1/4
c.	3/8
d.	1/2
e.	3/4

A brown-eyed couple already has a child with blue eyes. What is the probability that their next child will have blue eyes, assuming that brown eyes is dominant and blue eyes is recessive?

a.	0
b.	1/4
c.	1/2
d.	3/4
e.	1

A couple has already had 3 girls with cystic fibrosis, and were hoping to have a normal child for their fourth. What are the chances that the fourth child will be a normal male?

a.	0
b.	1/8
c.	1/4
d.	3/8
e.	1/2

A brown-eyed couple heterozygous for eye color are planning on having two children. What is the probability that both children will have blue eyes, assuming brown eyes is dominant and blue eyes is recessive?

a.	0
b.	1/32
c.	1/16
d.	1/4
e.	1/2

Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the probability that they will have a child with purple eyes?

a.	1/1
b.	1/2
c.	1/4
d.	3/4
e.	0

Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What are the chances that the alien couple will have a child with red eyes?

a.	1
b.	1/2
c.	1/4
d.	3/4
e.	0

Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the probability that the alien couple will have a child that is heterozygous for eye color?

a.	1
b.	1/2
c.	1/4
d.	3/4
e.	0

The principle of independent assortment is not true for:

a.	incomplete dominance.
b.	mutations.
c.	heterozygotes.
d.	homozygotes.
e.	X-linked genes.

The probability that two genes will be separated by crossing-over is related to:

a.	the phenotype that they control.
b.	how far the two genes are from the centromere.
c.	the distance between the two genes on the chromosome.
d.	whether the two genes are located on a sex chromosome.
e.	how far the genes are from the kinetochore.

Genes that tend to be inherited together are said to be:

a.	associated.
b.	related.
c.	similar.
d.	linked.
e.	alleles.

A ______________ is best used to demonstrate the linkage of two genes.

a.	monohybrid cross
b.	dihybrid cross
c.	monohybrid test cross
d.	two-allele test cross
e.	two-point test cross

In genetics, map units express the distance between:

a.	chromosomes during metaphase.
b.	two loci on a chromosome.
c.	alleles.
d.	polar bodies.
e.	homologous chromosomes.

In a two-point test cross, 36 of the offspring were recombinant types. The remaining 64 offspring were parental types. How many map units separate the two loci?

a.	28
b.	36
c.	64
d.	78
e.	100

The sex of most mammals, birds, and insects is determined by:

a.	the temperature.
b.	the environment.
c.	sex chromosomes.
d.	chance.
e.	None of the above.

The sex of a human is determined by:

a.	the number of chromosomes.
b.	the number of autosomes.
c.	the presence of only one X chromosome.
d.	the number of sex chromosomes.
e.	the presence of Y chromosome.

The offspring of two heterozygous gray-bodied, normal-winged flies should be 50% gray-bodied/normal wings (BbRr) and 50% black-bodied/vestigial wings (bbrr) because these alleles are linked. If a small number, say 15%, of the offspring are instead black-bodied with normal wings, this is most likely the result of:

a.	crossing-over.
b.	incomplete dominance.
c.	codominance.
d.	an error in meiosis.
e.	mutation.

An organism with the genotype of AaXx can produce gametes containing _________ if the two genes are unlinked.

a.	either Aa or Xx
b.	either AX, Ax, aX, ax
c.	AaXx
d.	AX or ax
e.	None of the above.

A lizard with a striped tail is crossed with one having a spotted head, producing normal looking (no stripes or spots) lizard progeny. What progeny would be expected to be produced by mating these progeny with each other, if the genes conferring stripes and spots were on different chromosomes?

a.	equal numbers of normal, striped, spotted and striped and spotted
b.	3 striped : 1 spotted
c.	9 striped and spotted : 3 spotted : 3 striped : 1 normal
d.	9 striped : 3 spotted : 1 striped or spotted
e.	9 normal : 3 striped : 3 spotted : 1 striped and spotted

A lizard with striped tails is crossed with one having a spotted head, producing normal looking (no stripes or spots) progeny. What progeny would be expected to be produced by mating one of these lizards with another that had a striped tail and spotted head, if the genes conferring stripes and spots were close together on the same chromosome?

a.	equal numbers of normal, striped, spotted and striped and spotted
b.	3 striped : 1 spotted
c.	mostly progeny that are striped or spotted
d.	mostly progeny that are normal or striped and spotted
e.	9 normal : 3 striped : 3 spotted : 1 striped and spotted

Why is color-blindness more common in males than in females?

a.	Because females would have to receive two copies of the recessive color blindness gene to actually express the trait.
b.	Because a male only needs to receive the recessive gene from his mother to be color-blind.
c.	Because color-blindness is an X-linked trait.
d.	All of the above.
e.	None of the above.

What are the possible phenotypes of the children if the mother has Type A blood and the father has type AB blood? (Use the Punnett square to verify your answer.)

 

a.	all AB
b.	A, B
c.	A, AB
d.	A, B, AB
e.	A, B, O

What are the possible phenotypes of the children if the mother's genotype is IAi for blood type and the father is IBi? (Use the Punnett square to verify your answer.)

 

a.	all AB
b.	A, B
c.	A, AB
d.	A, B, O
e.	A, B, AB, O

What are the predicted phenotypes of the male children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

 

a.	all normal
b.	3 normal : 1 hemophilia
c.	1 hemophilia : 1 normal
d.	1 hemophilia : 3 normal
e.	all hemophiliacs

What are the predicted phenotypes of the female children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.)

 

a.	all carriers
b.	3 homozygous normal : 1 carrier
c.	1 homozygous normal : 1 carrier
d.	1 hemophilia: 2 homozygous normal : 1 carrier
e.	all hemophiliacs

What are the possible genotypes of a female child from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics?

a.	XHXH or XHXh
b.	XHXh
c.	XHYH
d.	Hh
e.	HH

A Barr body in a mammalian female cell represents:

a.	an inactivated oocyte.
b.	a polar body.
c.	a degenerate nucleus.
d.	an inactivated X chromosome.
e.	an inactivated Y chromosome.

Calico cats are never male because:

a.	recessive genes are not usually expressed on their X chromosome.
b.	male hormones prevent expression of the calico phenotype.
c.	two different X chromosomes are needed for the expression of the calico phenotype.
d.	two different Y chromosomes are needed for the expression of the calico phenotype.
e.	one X chromosome and one Y chromosome are needed for the expression of the calico phenotype.

A diploid individual has a maximum of ____________ different alleles for a particular locus.

a.	one
b.	two
c.	three
d.	four
e.	more than four

A particular gene that controls seed coat color in peas also determines the susceptibility of these peas to a particular disease. This situation is referred to as:

a.	variegation.
b.	additive dominance.
c.	codominance.
d.	pleiotropy.
e.	incomplete dominance.

When certain medium height hybrid plants were crossed, they produced offspring that were dwarf, medium, and tall in a ratio of 1 : 2 : 1. This is an example of:

a.	variegation.
b.	hybrid vigor.
c.	incomplete dominance.
d.	epistasis.
e.	a polygenic trait.

Breeding a yellow dog with a brown dog produced puppies with both yellow and brown hairs intermixed. This is an example of:

a.	variegation.
b.	codominance.
c.	incomplete dominance.
d.	epistasis.
e.	a polygenic trait.

A gene that affects, prevents, or masks the expression of a gene at another locus is a(n) _________ gene.

a.	recessive
b.	dominant
c.	epistatic
d.	codominant
e.	plieotropic

The range of phenotypic possibilities that can develop from a certain dog genotype under different environmental conditions is called the:

a.	epistatic interaction.
b.	norm of reaction.
c.	nurture limit.
d.	genotype range.
e.	maximum phenotype.

________________________ refers to multiple independent pairs of genes having similar and additive effects on the same characteristic.

a.	Codominance
b.	Epistasis
c.	Polygenic inheritance
d.	Complete dominance
e.	Additive dominance

In the experiments of Griffith, the conversion of non-lethal R-strain bacteria to lethal S-strain bacteria:

a.	was the result of genetic mutation.
b.	was an example of the genetic exchange known as transformation.
c.	supported the case for proteins as the genetic material.
d.	could not be reproduced by other researchers.
e.	was an example of conjugation.

The first experimenters to use Griffith's transformation assay to identify the genetic material were:

a.	Meselson and Stahl.
b.	Watson and Crick.
c.	Franklin and Wilkins.
d.	Avery, MacLeod, and McCarty.
e.	Hershey and Chase.

The bacteriophages used in Alfred Hershey's and Martha Chase's experiments showed that:

a.	DNA was injected into bacteria.
b.	DNA and protein were injected into bacteria.
c.	DNA remained on the outer coat of bacteria.
d.	proteins were injected into bacteria.
e.	proteins were responsible for the production of new viruses within the bacteria.

The main reason scientists thought that proteins, rather than DNA, were the carriers of genetic material in the cell was:

a.	their presence within the nucleus.
b.	their abundance within the cell.
c.	the large number of possible amino acid combinations.
d.	their ability to self replicate within the cytoplasm.
e.	their ability to be exported from the cell.

Which of the following statements about DNA is false?

a.	DNA is capable of forming many different sequences.
b.	DNA contains thymine instead of uracil.
c.	DNA is double-stranded rather than single-stranded.
d.	DNA is only found in eukaryotic cells.
e.	DNA contains the sugar deoxyribose.

The information carried by DNA is incorporated in a code specified by the:

a.	phosphodiester bonds of the DNA strand.
b.	number of separate strands of DNA.
c.	size of a particular chromosome.
d.	specific nucleotide sequence of the DNA molecule.
e.	number of bases in a DNA strand.

X-ray diffraction studies are used to determine:

a.	the sequence of amino acids in protein molecules.
b.	the sequence of nucleic acids in nucleic acid molecules.
c.	the distances between atoms of molecules.
d.	the type of chemical under investigation.
e.	the wavelength of light emitted by chemicals.

The two molecules that alternate to form the backbone of a polynucleotide chain are:

a.	adenine and thymine.
b.	cytosine and guanine.
c.	sugar and phosphate.
d.	base and sugar.
e.	base and phosphate.

______________________ used x-ray diffraction to provide images of DNA.

a.	Watson and Crick
b.	Crick and Wilkins
c.	Franklin
d.	Franklin and Crick
e.	Watson and Wilkins

______________________ determined the structure of the molecule DNA.

a.	Crick and Wilkins
b.	Watson and Crick
c.	Franklin and Crick
d.	Franklin
e.	Watson, Crick, and Wilkins

Chargaff determined that DNA from any source contains about the same amount of guanine as __________.

a.	uracil
b.	thymine
c.	adenine
d.	cytosine
e.	guanine

X-ray crystallography showed that DNA:

a.	had the bases in the center of the molecule.
b.	had the sugars and phosphates on the outside of the molecule.
c.	was a very long molecule.
d.	was made of 2 strands.
e.	was a helix.

Why is DNA able to store large amounts of information?

a.	It contains a large number of different nucleotides.
b.	Its nucleotides can be arranged in a large number of possible sequences.
c.	It is capable of assuming a wide variety of shapes.
d.	The sugar and phosphates can be arranged in many different sequences.
e.	The nucleotides can be altered to form many different letters in the sequence.

In DNA, the genetic information:

a.	is contained in each strand.
b.	is contained in only one of the strands.
c.	is contained in both strands together.
d.	is contained in the differences between the 2 strands.
e.	None of the above.

Two chains of DNA must run in ____________ direction(s) and must be ____________ if they are to bond with each other.

a.	the same; uncomplementary
b.	opposite; uncomplementary
c.	parallel; uncomplementary
d.	parallel; complementary
e.	antiparallel; complementary

Hydrogen bonds can form between guanine and ____________, and between adenine and ____________.

a.	phosphate; sugar
b.	thymine; cytosine
c.	cytosine; thymine
d.	sugar; phosphate
e.	adenine; guanine

Figure 11-01
Use the figure below to answer the corresponding questions.

The portion of the molecule in box 5 of Figure 11-01 is:

a.	a hydrogen bond.
b.	a phosphate.
c.	a nucleotide.
d.	a pyrimidine.
e.	a protein.

In Figure 11-01, the portion of the molecule in box _________ is a pyrimidine.

a.	1
b.	3
c.	4
d.	1 and 3
e.	3 and 4

The portion of the molecule in box 3 of Figure 11-01 is:

a.	a sugar.
b.	a protein.
c.	a pyrimidine.
d.	a purine.
e.	a nucleotide.

Which of the following nucleotide sequences represents the complement to the DNA strand 5´ - AGATCCG- 3´?

a.	5´ – AGATCCG- 3´
b.	3´ – AGATCCG- 5´
c.	5´ – CTCGAAT- 3´
d.	3´ – CTCGAAT- 5´
e.	3´ – TCTAGGC- 5´

How is a single strand of DNA able to serve as a template for the synthesis of another strand?

a.	Nucleotides pair with those of the original strand to form a new strand.
b.	Hydrogen bonds holding the two strands together are easy to break, allowing one strand to be a template.
c.	A single strand of DNA is not able to serve as a template.
d.	One strand of DNA directs the synthesis of a new strand on its partner.
e.	Both A and B.

Which of the following best describes semiconservative replication?

a.	The translation of a DNA molecule into a complementary strand of RNA.
b.	A DNA molecule consists of one parental strand and one new strand.
c.	The number of DNA molecules is doubled with every other replication.
d.	The replication of DNA never takes place with 100% accuracy.
e.	The replication of DNA takes place at a defined period in the cell cycle.

Who first confirmed that the replication of DNA was semiconservative?

a.	Chargaff and Hershey
b.	Watson and Crick
c.	Avery and Griffith
d.	Meselson and Stahl
e.	Watson, Crick, and Wilkins

If DNA replication rejoined the 2 parental strands, it would be termed:

a.	dispersive.
b.	gradient.
c.	semiconservative.
d.	parental.
e.	conservative.

Meselson and Stahl separated DNA from different generations using:

a.	density gradient centrifugation.
b.	gel electrophoresis.
c.	an electron microscope.
d.	differential radioisotope labeling.
e.	None of the above.

When a DNA molecule containing a wrong base at one location in one strand is replicated:

a.	the mutation is corrected by the DNA polymerase enzyme.
b.	the mutation is ignored by the DNA polymerase enzyme.
c.	the mutation is copied into one of the two daughter molecules.
d.	the mutation is copied into both of the daughter molecules.
e.	the replication is stopped.

What prevents knot formation in replicating DNA?

a.	protosomes
b.	topoisomerases
c.	scaffolding proteins
d.	chromatin
e.	histones

Why does DNA synthesis only proceed in the 5´to 3´ direction?

a.	Because DNA polymerases can only add nucleotides to the 3´ end of a polynucleotide strand.
b.	Because the 3´ end of the polynucleotide molecule is more electronegative than the 5´ end.
c.	Because that is the direction in which the two strands of DNA unzip.
d.	Because that is the only direction that the polymerase can be oriented.
e.	Because the chromosomes are always aligned in the 5´ to 3´ direction in the nucleus.

How is the chromosome of a bacterial cell replicated?

a.	The linear DNA molecule is replicated from multiple origins of replication bidirectionally.
b.	The linear DNA molecule is replicated from one origin of replication bidirectionally.
c.	The circular DNA molecule is replicated from multiple origins of replication bidirectionally.
d.	The circular DNA molecule is replicated from one origin of replication bidirectionally.
e.	The circular DNA molecule is replicated from one origin of replication unidirectionally.

How are the chromosomes of a eukaryote cell replicated?

a.	The linear DNA molecules are replicated from multiple origins of replication bidirectionally.
b.	The linear DNA molecules are replicated from one origin of replication bidirectionally.
c.	The circular DNA molecules are replicated from multiple origins of replication bidirectionally.
d.	The circular DNA molecules are replicated from one origin of replication bidirectionally.
e.	The linear DNA molecules are replicated from one origin of replication unidirectionally.

Which of the following adds new nucleotides to a growing DNA chain?

a.	DNA polymerase
b.	DNA helicase
c.	RNA primer
d.	primase
e.	RNA polymerase

Which of the following cause the unwinding of the DNA double helix?

a.	DNA polymerase
b.	DNA helicase
c.	RNA primer
d.	primosome
e.	RNA polymerase

In DNA replication, the lagging strand:

a.	is synthesized as a series of Okazaki fragments.
b.	is synthesized as a complementary copy of the leading strand.
c.	pairs with the leading strand by complementary base pairing.
d.	is made up entirely of RNA primers.
e.	is not synthesized until the synthesis of the leading strand is completed.

Figure 11-02
Use the figure below to answer the corresponding questions.

The correct designation for the DNA strand labeled C in Figure 11-02 is:

a.	the leading strand.
b.	3´.
c.	Okazaki fragments.
d.	polymerase.
e.	None of the above.

The segments labeled F in Figure 11-02 are responsible for:

a.	linking short DNA segments.
b.	synthesizing the leading strand.
c.	forming the replication fork.
d.	initiating DNA synthesis.
e.	unwinding the DNA double helix.

The enzyme represented by the letter D in Figure 11-02 is responsible for:

a.	linking short DNA segments.
b.	synthesizing the leading strand.
c.	forming the replication fork.
d.	forming nucleosomes.
e.	unwinding the DNA double helix.

The structures represented by the letter E in Figure 11-02 are called:

a.	leading fragments
b.	Okazaki fragments.
c.	replication forks.
d.	nucleosomes.
e.	DNA polymerases.

In replication, once the DNA strands have been separated, reformation of the double helix is prevented by:

a.	DNA helicase enzyme.
b.	helix-destabilizing proteins.
c.	DNA polymerases.
d.	ATP.
e.	GTP.

Enzymes called ____________ form nicks in the DNA molecules to prevent the formation of knots in the DNA helix during replication.

a.	topoisomerases
b.	helix-destabilizing enzymes
c.	DNA polymerases
d.	RNA polymerases
e.	DNA ligases

The DNA strand that is replicated smoothly and continuously is called the:

a.	primary strand.
b.	first strand.
c.	leading strand.
d.	alpha strand.
e.	lagging strand.

The final product of DNA replication is:

a.	mRNA, tRNA, and rRNA molecules.
b.	a wide variety of proteins.
c.	DNA fragments.
d.	two DNA molecules, each of which contains one new and one old DNA strand.
e.	the enzymes needed for further processes, such as DNA polymerase.

A replication fork:

a.	is only seen in prokaryotic chromosomes.
b.	is only seen in bacterial cells.
c.	is a Y-shaped structure where both DNA strands are replicated simultaneously.
d.	is a site where one DNA strand serves as a template, but the other strand is not replicated.
e.	is created by the action of the enzyme RNA polymerase.

The 5´ end of each Okazaki fragment begins with:

a.	the same RNA primer that began synthesis on the leading strand.
b.	a DNA primer binding to the template DNA.
c.	DNA polymerase binding to the template DNA.
d.	a separate RNA primer.
e.	a small DNA primer.

Okazaki fragments are joined together by:

a.	RNA polymerase.
b.	DNA ligase.
c.	DNA polymerase.
d.	RNA ligase.
e.	primase.

Primase is the enzyme responsible for:

a.	unwinding the DNA double strand to allow DNA polymerase access to the template DNA.
b.	introducing nicks into the DNA double strand in order to prevent the formation of knots.
c.	hydrolyzing ATP to facilitate DNA unwinding.
d.	making short strands of RNA at the site of replication initiation.
e.	forming a replication fork in the DNA double helix.

____________, the ends of eukaryotic chromosomes, shorten with every cell replication event.

a.	Centromeres
b.	Telomeres
c.	Kinetochores
d.	Primosomes
e.	Nucleosomes

The ends of eukaryotic chromosomes can be lengthened by:

a.	apoptosis.
b.	reverse transcriptase.
c.	primase.
d.	telomerase.
e.	DNA polymerase.

Cancer cells differ from noncancerous cells in that:

a.	they have elevated levels of telomerase.
b.	they are virtually immortal.
c.	they have the ability to resist apoptosis.
d.	they can maintain telomere length as they divide.
e.	All of the above.

In 1998, Bodnar and her colleagues found that by introducing genes coding for telomeres into cultured human cells:

a.	the cells underwent more cell divisions than normal.
b.	the cells underwent fewer cell divisions than normal.
c.	the cells all lived indefinitely.
d.	the cells underwent gene expression more vigorously.
e.	the cell cycle shortened.

Which of the following is not a reason why Neurospora is an ideal organism to study the effects of genetic mutations?

a.	Neurospora is easy to grow.
b.	Neurospora grows as a haploid organism.
c.	Neurospora is easy to genetically manipulate.
d.	Neurospora reproduces both sexually and asexually.
e.	Neurospora contains homologous chromosomes that are easily viewed with a light microscope.

Garrod first proposed that:

a.	metabolic defects were due to a lack of an enzyme.
b.	metabolic defects were due to excess enzyme.
c.	metabolic defects were due to chromosomal changes.
d.	mutations were inheritable.
e.	metabolic defects did not occur in humans.

Why was it important in the studies of Beadle and Tatum that Neurospora is haploid?

a.	Because it is easier to grow haploid molds in the laboratory.
b.	Because haploid molds have simpler nutritional requirements than do diploid molds.
c.	Because a mutation that arises is not masked by a normal allele on a homologous chromosome.
d.	Because haploid Neurospora will always mutate.
e.	Because diploid Neurospora will always mutate.

Beadle and Tatum began their studies with wild-type Neurospora, which is:

a.	Neurospora that only grows in the wild.
b.	a mutant strain that will only grow in the lab on complete medium.
c.	a strain that will not grow in the lab.
d.	a virulent strain of Neurospora.
e.	a normal phenotype that will grow on minimal medium.

Beadle and Tatum irradiated Neurospora and initially grew the mutant strains on complete medium. How were they able to determine what type of mutation each strain had?

a.	By growing the mold on a complete medium with extra vitamins and nutrients.
b.	By growing the mold on minimal media supplemented with different combinations of amino acids, vitamins, etc.
c.	By growing the mold in its diploid form to see which traits were masked.
d.	By comparing Neurospora to other species of mold.
e.	By observing the marked differences in morphology between the different strains.

What conclusions did Beadle and Tatum reach with their studies of Neurospora?

a.	Each mutant gene affected several enzymes.
b.	Each mutant gene affected a pair of enzymes.
c.	Each mutant gene affected only one enzyme.
d.	Mutant genes had no effect on the enzymes produced by the cells.
e.	None of the above.

Linus Pauling demonstrated that:

a.	the structure of hemoglobin was altered by a mutation of a single gene.
b.	mutations only caused defects in enzymes.
c.	mutations alter the structure of RNA, but not proteins.
d.	mutations were inherited.
e.	the structure of hemoglobin was altered by mutations in any of a dozen genes.

Figure 12-01
Use the figure below to answer the corresponding questions.

The experimental design in Figure 12-01 was used to examine:

a.	the relationship between genetic changes and metabolic enzymes.
b.	the mutation rate of Neurospora.
c.	resistance of Neurospora to genetic poisons.
d.	toxicity of arginine metabolites.
e.	growth of Neurospora in the presence of different antibiotics.

The conclusion associated with mutant strain III in Figure 12-01 was that:

a.	it contained all the enzymes needed for normal metabolism.
b.	it was missing all the enzymes for metabolism of amino acids.
c.	it was missing an enzyme for metabolism and could not synthesize arginine.
d.	it was missing an enzyme for metabolism and could not synthesize citrulline.
e.	it was missing an enzyme for metabolism and could not synthesize ornithine.

One of the mRNA codons specifying the amino acid leucine is 5´-CUA-3´. Its corresponding anticodon is:

a.	5´-GAT-3´.
b.	3´-AUC-5´.
c.	3´-GAU-5´.
d.	3´-GAT-5´.
e.	5´-GAU-3´.

Which of the following is a characteristic of uracil?

a.	The ability to bond with adenine.
b.	The ability to bond with guanine.
c.	It is a purine.
d.	The ability to bond with cytosine.
e.	It contains two nitrogenous rings.

RNA differs from DNA in all the following except:

a.	RNA is single stranded and DNA is double stranded.
b.	RNA is a larger molecule than DNA.
c.	RNA contains uracil and DNA contains thymine.
d.	RNA contains ribose and DNA contains deoxyribose.
e.	None of the above.

Ribose differs from deoxyribose by having:

a.	two attached bases.
b.	one less oxygen.
c.	an extra hydroxyl group.
d.	an extra carbon in the ring.
e.	None of the above.

Uracil forms a complementary pair with ____________ in RNA and _____________ in DNA.

a.	adenine; adenine
b.	adenine; thymine
c.	thymine; thymine
d.	uracil; adenine
e.	adenine; uracil

RNA synthesis is also known as:

a.	elongation.
b.	reverse transcription.
c.	termination.
d.	translation.
e.	transcription.

All RNA except for _________ is made from DNA.

a.	tRNA
b.	mRNA
c.	rRNA
d.	snRNA
e.	None of the above.

The total number of different three-base combinations of the four nucleic acid bases is:

a.	12.
b.	16.
c.	20.
d.	64.
e.	256.

During protein synthesis, ribosomes:

a.	attach to the mRNA molecule and travel along its length.
b.	attach to the DNA molecule and travel along its length to produce an mRNA molecule.
c.	translate mRNA into tRNA.
d.	transcribe mRNA to tRNA.
e.	translate mRNA into DNA.

How is the four-letter language of nucleic acids converted into the 20-word language of amino acids?

a.	The 4 nucleic acid bases combine in 2-letter combinations that define different amino acids.
b.	The 4 nucleic acid bases combine in 3-letter sequences that define different amino acids.
c.	Triplets of the 2-letter nucleic acid bases are translated into the 20 different amino acids.
d.	The 4 bases each specify 1 amino acid, which give rise to the remaining 16 amino acids.
e.	The 4 bases are first converted into tRNA molecules, which can each attach to 5 amino acids.

Why is only one strand of DNA transcribed into mRNA?

a.	Because mRNA is only required in small quantities.
b.	Because transcribing both DNA strands would produce different amino acid sequences.
c.	Because the other strand would produce the same amino acid sequence in reverse order.
d.	Because all genes are located on the same DNA strand, while the other strand acts as protection.
e.	Because the other strand is transcribed directly into amino acids.

Initiation of transcription requires:

a.	a promoter sequence.
b.	DNA polymerase.
c.	an RNA primer.
d.	a DNA primer.
e.	Okazaki fragments.

A sequence of bases located upstream from a reference point occurs:

a.	towards the 3´ end of the amino acid sequence.
b.	towards the 5´ end of the mRNA sequence.
c.	towards the 3´ end of the mRNA sequence.
d.	towards the 5´ end of the transcribed DNA strand.
e.	towards the carboxyl end of the amino acid sequence.

How does the first nucleotide at the 5´ end of a new mRNA chain differ from the other nucleotides in the chain?

a.	The first nucleotide is always a uracil.
b.	The first nucleotide is always a cytosine.
c.	The first nucleotide retains its triphosphate group, while the others do not.
d.	The first nucleotide does not retain its triphosphate group, while the others in the chain do.
e.	The first nucleotide is always a modified cytosine.

Figure 12-02
Use the figure below to answer the corresponding questions.

In Figure 12-02, the transcription process begins at the area labeled:

a.	A.
b.	D.
c.	E.
d.	G.
e.	None of the above.

The component labeled B in Figure 12-02 is:

a.	DNase.
b.	DNA polymerase.
c.	RNA primase.
d.	RNA polymerase.
e.	reverse transcriptase.

The transcript in Figure 12-02 is labeled:

a.	D.
b.	C.
c.	G.
d.	H.
e.	A.

The process illustrated in Figure 12-02 is:

a.	DNA synthesis.
b.	translation.
c.	transcription.
d.	a frame shift mutation.
e.	protein synthesis.

Leader sequences contain signals that:

a.	prevent enzymes from degrading the newly synthesized mRNA.
b.	inhibit ribosome binding until the appropriate time.
c.	initiate chain termination.
d.	allow the ribosomes to be properly positioned to translate the message.
e.	allow tRNA molecules to successfully bind to mRNA.

Aminoacyl-tRNA synthetases ________ link ________ to their respective tRNA molecules.

a.	ionically; mRNAs
b.	loosely; mRNAs
c.	terminally; codons
d.	covalently; amino acids
e.	enzymatically; codons

Which of the following numbered terms represents the correct order of sequences in a prokaryotic mRNA molecule as it was synthesized?

1.	3´ trailing sequences
2.	coding sequences
3.	leader sequences
4.	termination signals

a.	1 ® 2 ® 3 ® 4
b.	3 ® 2 ® 4 ® 1
c.	2 ® 1 ® 4 ® 3
d.	4 ® 2 ® 1 ® 3
e.	3 ® 4 ® 2 ® 1

Which of the following serves as an "adapter" in protein synthesis and bridges the gap between mRNA and proteins?

a.	tRNA
b.	cDNA
c.	rRNA
d.	promoter sequences
e.	DNA

The codon is found in the:

a.	template strand of DNA.
b.	non-template strand of DNA.
c.	mRNA.
d.	tRNA.
e.	rRNA.

The tRNA:

a.	must be recognized by ribosomes.
b.	must have an anticodon.
c.	must have an attachment site for the amino acid.
d.	must be recognized by a specific aminoacyl-tRNA synthetase that adds the correct amino acid.
e.	All of the above.

Which of the following numbered events represents the correct sequence of events of prokaryotic translation initiation?

1.	large ribosomal subunit binds to initiation complex
2.	initiation tRNA binds small ribosomal subunit
3.	initiation complex binds to ribosome recognition sequence on mRNA

a.	1 ® 2 ® 3
b.	1 ® 3 ® 2
c.	2 ® 1 ® 3
d.	2 ® 3 ® 1
e.	3 ® 2 ® 1

Where is the amino-acid binding site located on the tRNA molecule?

a.	in the middle of the loop
b.	at the end of a "stem" that is the 3´ end of the molecule
c.	in the first loop
d.	along the longest stretch of base pairing in the molecule
e.	on the 5´ end of the molecule

Figure 12-03
Use the figure below to answer the corresponding questions.

In Figure 12-03, the portion of the molecule in the figure that contains the anti-codon is:

a.	1.
b.	2.
c.	3.
d.	4.
e.	5.

In Figure 12-03, the portion of the molecule labeled 5 is:

a.	the attached amino acid.
b.	a double-stranded region.
c.	a single-stranded region.
d.	the anti-codon.
e.	the codon.

The enzyme peptidyl transferase, which catalyzes the transfer of the polypeptide chain attached to the tRNA in the ____________ site to the aminoacyl-tRNA in the ____________ site, is thought to be a(an) ____________ molecule and not a protein.

a.	A; P; rDNA
b.	P; A; tRNA
c.	A; P; mRNA
d.	P; A; rRNA
e.	P; A; sugar

Translocation is the process whereby the __________ moves in order to place the tRNA bound to the growing polypeptide chain in the __________ site, thereby freeing the __________ site for a new aminoacyl-tRNA.

a.	mRNA; A; P
b.	ribosome; P; A
c.	tRNA; P; A
d.	ribosome; A; P
e.	tRNA; A; P

Following peptide bond formation between the amino acid in the A site on the ribosome and the growing polypeptide chain, the tRNA in the A site:

a.	releases the growing polypeptide chain.
b.	picks up another amino acid to add to the chain.
c.	moves to the P site of the ribosome.
d.	forms a peptide bond with A site of the ribosome.
e.	forms a covalent bond with the P site of the ribosome.

If a human gene mRNA were placed into a cell of yeast, it would be:

a.	degraded immediately.
b.	translated into a repeating amino acid chain.
c.	translated into a chain of random amino acids not resembling the protein in humans.
d.	translated into the protein that is found in humans.
e.	integrated into the genome of the yeast.

In all organisms, the AUG codon codes for:

a.	the initiation of translation.
b.	the termination of transcription.
c.	the termination of chain elongation.
d.	the amino acid valine.
e.	a termination tRNA molecule.

A polyribosomes is:

a.	a complex of many ribosome and an mRNA.
b.	a complex of many ribosomes in eukaryotes.
c.	an initiation complex in eukaryotes.
d.	an elongation complex in eukaryotes.
e.	a complex of a ribosome with its two subunits and several mRNAs.

Introns in pre mRNA are known to:

a.	code for specific protein domains.
b.	undergo excision, whereby they are spliced out of the message.
c.	be able to move within the mRNA, thereby giving rise to new exon combinations.
d.	protect pre mRNA from enzyme degradation.
e.	code for important amino acid sequences.

Proteins synthesized in E. coli have which of the following at their amino terminal end?

a.	N-formyl-methionine
b.	N-acetyl-adenine
c.	adenine triphosphate
d.	the AUG codon
e.	the UUU codon

The wobble hypothesis states that:

a.	more than one ribosome can bind to an mRNA molecule.
b.	some amino acids are coded for by more than one codon.
c.	there is more than one stop codon in the genetic code.
d.	a particular amino acid may be linked to more than one type of tRNA molecule.
e.	certain tRNA anticodons can pair with more than one codon sequence.

Binding of the appropriate aminoacyl-tRNA to the A site requires:

a.	no additional energy.
b.	the input of two ATP molecules to supply the needed energy.
c.	energy supplied by GTP.
d.	activation of the A site.
e.	phosphorylation of the tRNA molecule.

Translocation in translation requires:

a.	no additional energy.
b.	activation of the P site.
c.	the input of two ATP molecules to supply the needed energy.
d.	energy supplied by GTP.
e.	phosphorylation of the mRNA molecule.

Interrupted coding sequences include long sequences of bases that do not code for amino acids. These noncoding sequences, called ____________, are found in ____________ cells.

a.	exons; prokaryotic
b.	introns; prokaryotic
c.	exons; eukaryotic
d.	introns; eukaryotic
e.	None of the above.

An mRNA "5´ cap":

a.	prevents translation.
b.	facilitates binding of ribosomes.
c.	marks the mRNA for degradation.
d.	decreases the half-life of the mRNA.
e.	protects newly synthesized mRNA from degradation.

The 3´ end of eukaryotic pre-mRNAs are changed by:

a.	removing the last phosphate group.
b.	adding a "cap."
c.	copying the last few bases so that it can form a duplex structure.
d.	cutting and adding 100-250 adenine nucleotides.
e.	phosphorylation of the mRNA molecule.

Walter Gilbert proposed that exons are:

a.	remnants of older life forms.
b.	sequences that code for protein domains that are shuffled to form new proteins.
c.	the result of mutation of introns.
d.	not present in prokaryotes.
e.	sequences that interrupt the coding sequences of proteins.

Retroviruses or RNA tumor viruses use __________ to make DNA:

a.	DNA polymerase
b.	DNA-dependent RNA polymerase
c.	RNA polymerase
d.	primase
e.	reverse transcriptase

Substitution of one base pair for another can result in a ____________ mutation that results in the conversion of an amino acid specifying codon to a termination codon.

a.	nonsense
b.	frameshift
c.	chromosomal
d.	missense
e.	None of the above.

A mutation that replaces one amino acid in a protein with another is called a ____________ mutation.

a.	frameshift
b.	recombinant
c.	nonsense
d.	missense
e.	neutral

Frameshift mutations result from:

a.	the substitution of one base pair for another.
b.	the substitution of more than one base pair.
c.	the insertion or deletion of one or two base pairs.
d.	the substitution of a stop codon for an amino acid-specifying codon.
e.	the substitution of a start codon for an amino acid codon.

A gene can now be defined as:

a.	a DNA sequence that carries information to produce a specific RNA or protein product.
b.	a DNA nucleotide sequence that carries information to produce a specific polypeptide.
c.	a DNA or RNA sequence that carries information to produce a single polypeptide.
d.	a DNA nucleotide sequence that carries information to produce an enzyme.
e.	a DNA or RNA sequence that carries information to produce a specific polypeptide.