Life 120: Exam 3 Questions Flashcards

Starting with a fertilized egg (zygote), a series of five cell divisions would produce an early embryo with how many cells?
A) 4
B) 8

C) 16
D) 32
E) 64

If there are 20 chromatids in a cell, how many centromeres are there?

A) 10
B) 20
C) 30

D) 40
E) 80

In eukaryotic cells, chromosomes are composed of DNA

A) and RNA.
B) only.
C) and proteins.

D) and phospholipids.

What is produced if a cell divides by mitosis but does not undergo cytokinesis?

A) two cells, one cell containing two nuclei and a second cell without a nucleus
B) two cells, each cell with half of the genetic material of the parent cell
C) one cell with one nucleus containing half of the genetic material of the parent cell

D) one cell with two nuclei, each identical to the nucleus of the parent cell

Humans produce skin cells by mitosis and gametes by meiosis. The nuclei of skin cells produced by mitosis will have
A) half as much DNA as the nuclei of gametes produced by meiosis.
B) the same amount of DNA as the nuclei of gametes produced by meiosis.

C) twice as much DNA as the nuclei of gametes produced by meiosis.
D) four times as much DNA as the nuclei of gametes produced by meiosis.

Compared to most prokaryotic cells, eukaryotic cells typically have

A) more DNA molecules and larger genomes.
B) the same number of DNA molecules but larger genomes.
C) the same number of DNA molecules and similarly sized genomes.

D) fewer DNA molecules but larger genomes.

E) fewer DNA molecules and smaller genomes.

At which phase are centrioles beginning to move apart in animal cells?

A) telophase
B) anaphase
C) prometaphase

D) metaphase
E) prophase

If there are 20 centromeres in a cell at anaphase, how many chromosomes are there in each daughter cell following cytokinesis?
A) 10
B) 20

C) 30
D) 40
E) 80

Where do the microtubules of the spindle originate during mitosis in animal cells?

A) centromere
B) centrosome
C) centriole

D) chromatid
E) kinetochore

Taxol is an anticancer drug extracted from the Pacific yew tree. In animal cells, Taxol disrupts microtubule formation by binding to microtubules and accelerating their assembly from the protein precursor tubulin. Surprisingly, this stops mitosis. Specifically, Taxol must affect
A) the formation of the mitotic spindle.
B) anaphase.
C) formation of the centrioles.
D) chromatid assembly.
E) the S phase of the cell cycle.

Which of the following are primarily responsible for cytokinesis in plant cells but not in animal cells?
A) kinetochores
B) Golgi-derived vesicles

C) actin and myosin
D) centrioles and centromeres
E) tubulin and dynein

Movement of the chromosomes during anaphase would be most affected by a drug that prevents
A) nuclear envelope breakdown.
B) cell wall formation.

C) elongation of microtubules.
D) shortening of microtubules.
E) formation of a cleavage furrow.

Measurements of the amount of DNA per nucleus were taken on a large number of cells from a growing fungus. The measured DNA levels ranged from 3 to 6 picograms per nucleus. In which stage of the cell cycle did the nucleus contain 6 picograms of DNA?
A) G0
B) G1
C) S
D) G2
E) M

A group of cells is assayed for DNA content immediately following mitosis and is found to have an average of 8 picograms of DNA per nucleus. How many picograms would be found at the end of S and the end of G2?
A) 8; 8

B) 8; 16
C) 16; 8
D) 16; 16
E) 12; 16

The beginning of anaphase is indicated by which of the following?

A) Chromatids lose their kinetochores.
B) Cohesin attaches the sister chromatids to each other.
C) Cohesin is cleaved enzymatically.

D) Kinetochores attach to the metaphase plate.

E) Spindle microtubules begin to polymerize.

During which phase of mitosis do the chromatids become chromosomes?

A) telophase
B) anaphase
C) prophase

D) metaphase
E) cytokinesis

What is a cleavage furrow?
A) a ring of vesicles forming a cell plate
B) the separation of divided prokaryotes
C) a groove in the plasma membrane between daughter nuclei
D) the metaphase plate where chromosomes attach to the spindle
E) the space that is created between two chromatids during anaphase

Using which of the following techniques would enable your lab group to distinguish between a cell in G2 and a cell from the same organism in G1?
A) fluorescence microscopy
B) electron microscopy

C) spectrophotometry
D) radioactive-labeled nucleotides

E) labeled kinetochore proteins

You have the technology necessary to measure each of the following in a sample of animal cells: chlorophylls, organelle density, picograms of DNA, cell wall components, and enzymatic activity. Which would you expect to increase significantly from M to G1?
A) organelle density and enzymatic activity

B) cell wall components and DNA

C) chlorophyll and cell walls
D) organelle density and cell walls

E) chlorophyll and DNA

A plant-derived protein known as colchicine can be used to poison cells by blocking the formation of the spindle. Which of the following would result if colchicine is added to a sample of cells in G2?
A) The cells would immediately die.
B) The cells would be unable to begin M and stay in G2.
C) The chromosomes would coil and shorten but have no spindle to which to attach.
D) The chromosomes would segregate but in a disorderly pattern.
E) Each resultant daughter cell would also be unable to form a spindle.

Motor proteins require which of the following to function in the movement of chromosomes toward the poles of the mitotic spindle?
A) intact centromeres
B) a microtubule-organizing center

C) a kinetochore attached to the metaphase plate

D) ATP as an energy source
E) synthesis of cohesin

When a cell is in late anaphase of mitosis, which of the following will we see?

A) a clear area in the center of the cell
B) chromosomes clustered at the poles
C) individual chromatids beginning to separate from one another

D) chromosomes clustered tightly at the center

E) breaking down of the nuclear envelope

Cells from advanced malignant tumors often have very abnormal chromosomes as well as an abnormal number of chromosomes. What might explain the association between malignant tumors and chromosomal abnormalities?
A) Cancer cells are no longer density dependent.

B) Cancer cells are no longer anchorage dependent.
C) Cell cycle checkpoints are not in place to stop cells with chromosome abnormalities.

D) Chromosomally abnormal cells still have normal metabolism.
E) Transformation introduces new chromosomes into cells.

Which is the first checkpoint in the cell cycle where a cell will be caused to exit the cycle if this point is not passed?
A) G0
B) G1

C) G2

D) S

E) previous M

Which of the following is released by platelets in the vicinity of an injury?

A) PDGF
B) MPF
C) protein kinase

D) cyclin
E) Cdk

Which of the following is a protein synthesized at specific times during the cell cycle that associates with a kinase to form a catalytically active complex?
A) PDGF
B) MPF

C) protein kinase
D) cyclin
E) Cdk

Which of the following is a protein maintained at constant levels throughout the cell cycle that requires cyclin to become catalytically active?
A) PDGF
B) MPF

C) protein kinase
D) cyclin
E) Cdk

Which of the following triggers the cell's passage past the G2 checkpoint into mitosis? A) PDGF

B) MPF
C) protein kinase
D) cyclin
E) Cdk

The cyclin component of MPF is destroyed toward the end of which phase?

A) G0
B) G1
C) S

D) G2

E) M

Proteins that are involved in the regulation of the cell cycle, and that show fluctuations in concentration during the cell cycle, are called
A) ATPases.
B) kinetochores.

C) kinases.
D) estrogen receptors.
E) cyclins.

The MPF protein complex turns itself off by
A) activating a process that destroys cyclin components.

B) activating an enzyme that stimulates cyclin.
C) binding to chromatin.
D) exiting the cell.
E) activating the anaphase-promoting complex.

Density-dependent inhibition is explained by which of the following?
A) As cells become more numerous, they begin to squeeze against each other, restricting their size and ability to produce control factors.
B) As cells become more numerous, the cell surface proteins of one cell contact the adjoining cells and they stop dividing.
C) As cells become more numerous, the protein kinases they produce begin to compete with each other, such that the proteins produced by one cell essentially cancel those produced by its neighbor.
D) As cells become more numerous, more and more of them enter the S phase of the cell cycle.

E) As cells become more numerous, the level of waste products increases, eventually slowing down metabolism.

Which of the following is true concerning cancer cells?
A) They do not exhibit density-dependent inhibition when growing in culture.
B) When they stop dividing, they do so at random points in the cell cycle.
C) They are not subject to cell cycle controls.
D) When they stop dividing, they do so at random points in the cell cycle, and they are not subject to cell cycle controls.
E) When they stop dividing, they do so at random points in the cell cycle; they are not subject to cell cycle controls; and they do not exhibit density-dependent inhibition when growing in culture.

Which of the following describes cyclin-dependent kinase (Cdk)?
A) Cdk is inactive, or "turned off," in the presence of cyclin.
B) Cdk is present throughout the cell cycle.
C) Cdk is an enzyme that attaches phosphate groups to other proteins.
D) Cdk is inactive, or "turned off," in the presence of cyclin and it is present throughout the cell cycle.

E) Cdk is present throughout the cell cycle and is an enzyme that attaches phosphate groups to other proteins.

Besides the ability of some cancer cells to overproliferate, what else could logically result in a tumor?
A) enhanced anchorage dependence
B) changes in the order of cell cycle stages

C) lack of appropriate cell death
D) inability to form spindles
E) inability of chromosomes to meet at the metaphase plate

Why do neurons and some other specialized cells divide infrequently?

A) They no longer have active nuclei.
B) They no longer carry receptors for signal molecules.
C) They have been shunted into G0.

D) They can no longer bind Cdk to cyclin.

E) They show a drop in MPF concentration.

Which of the following most accurately describes a cyclin?
A) It is present in similar concentrations throughout the cell cycle.

B) It is activated to phosphorylate by complexing with a Cdk.
C) It decreases in concentration when MPF activity increases.
D) It activates a Cdk molecule when it is in sufficient concentration.

E) It activates a Cdk when its concentration is decreased.

All cell cycle checkpoints are similar in which way?
A) They respond to the same cyclins.
B) They utilize the same Cdks.
C) They give the go-ahead signal to progress to the next checkpoint.

D) They each have only one cyclin/Cdk complex.

E) They activate or inactivate other proteins.

At the M phase checkpoint, the complex allows for what to occur?

A) Separase enzyme cleaves cohesins and allows chromatids to separate.

B) Cohesins alter separase to allow chromatids to separate.
C) Kinetochores are able to bind to spindle microtubules.
D) All microtubules are made to bind to kinetochores.
E) Daughter cells are allowed to pass into G1.

Anchorage dependence of animal cells in vitro or in vivo depends on which of the following?

A) attachment of spindle fibers to centrioles
B) response of the cell cycle controls to signals from the plasma membrane
C) the absence of an extracellular matrix

D) the binding of cell-surface phospholipids to those of adjoining cells

E) the binding of cell-surface phospholipids to the substrate

Researchers began a study of a cultured cell line. Their preliminary observations showed them that the cell line did not exhibit either density-dependent inhibition or anchorage dependence. What could they conclude right away?
A) The cells originated in the nervous system.

B) The cells are unable to form spindle microtubules.
C) The cells have altered series of cell cycle phases.
D) The cells show characteristics of tumors.
E) The cells were originally derived from an elderly organism.

For a chemotherapeutic drug to be useful for treating cancer cells, which of the following is most desirable?
A) It is safe enough to limit all apoptosis.
B) It does not alter metabolically active cells.

C) It only attacks cells that are density dependent.

D) It interferes with cells entering G0.
E) It interferes with rapidly dividing cells.

You have a series of cells, all of which were derived from tumors, and you first need to find out which ones are malignant. What could you do?
A) See which ones are not overproliferating.
B) Find out which ones have a higher rate of apoptosis.

C) Karyotype samples to look for unusual size and number of chromosomes.

D) Measure metastasis.
E) Time their cell cycles.

These protists are intermediate in what sense?
A) They reproduce by binary fission in their early stages of development and by mitosis when they are mature.
B) They never coil up their chromosomes when they are dividing.
C) They use mitotic division but only have circular chromosomes.
D) They maintain a nuclear envelope during division.
E) None of them form spindles.

What is the most probable hypothesis about these intermediate forms of cell division?
A) They represent a form of cell reproduction that must have evolved completely separately from those of other organisms.
B) They demonstrate that these species are not closely related to any of the other protists and may well be a different kingdom.
C) They rely on totally different proteins for the processes they undergo.
D) They may be more closely related to plant forms that also have unusual mitosis.
E) They show some but not all of the evolutionary steps toward complete mitosis.

Which of the following questions might be answered by such a method?

A) How many cells are produced by the culture per hour?
B) What is the length of the S phase of the cell cycle?
C) When is the S chromosome synthesized?

D) How many picograms of DNA are made per cell cycle?

E) When do spindle fibers attach to chromosomes?

The research team used the setup to study the incorporation of labeled nucleotides into a culture of lymphocytes and found that the lymphocytes incorporated the labeled nucleotide at a significantly higher level after a pathogen was introduced into the culture. They concluded that

A) the presence of the pathogen made the experiment too contaminated to trust the results.
B) their tissue culture methods needed to be relearned.
C) infection causes lymphocytes to divide more rapidly.
D) infection causes cell cultures in general to reproduce more rapidly.
E) infection causes lymphocyte cultures to skip some parts of the cell cycle.

Through a microscope, you can see a cell plate beginning to develop across the middle of a cell and nuclei forming on either side of the cell plate. This cell is most likely
A) an animal cell in the process of cytokinesis.
B) a plant cell in the process of cytokinesis.

C) an animal cell in the S phase of the cell cycle.

D) a bacterial cell dividing.
E) a plant cell in metaphase.

In the cells of some organisms, mitosis occurs without cytokinesis. This will result in A) cells with more than one nucleus.
B) cells that are unusually small.
C) cells lacking nuclei.

D) destruction of chromosomes.
E) cell cycles lacking an S phase.

Which of the following does not occur during mitosis?

A) condensation of the chromosomes
B) replication of the DNA
C) separation of sister chromatids

D) spindle formation
E) separation of the spindle poles

A particular cell has half as much DNA as some other cells in a mitotically active tissue. The cell in question is most likely in
A) G1.
B) G2.

C) prophase.
D) metaphase.
E) anaphase.

The drug cytochalasin B blocks the function of actin. Which of the following aspects of the animal cell cycle would be most disrupted by cytochalasin B?
A) spindle formation
B) spindle attachment to kinetochores

C) DNA synthesis
D) cell elongation during anaphase
E) cleavage furrow formation and cytokinesis

If a horticulturist breeding gardenias succeeds in having a single plant with a particularly desirable set of traits, which of the following would be her most probable and efficient route to establishing a line of such plants?
A) Backtrack through her previous experiments to obtain another plant with the same traits.

B) Breed this plant with another plant with much weaker traits.
C) Clone the plant asexually to produce an identical one.
D) Force the plant to self-pollinate to obtain an identical one.
E) Add nitrogen to the soil of the offspring of this plant so the desired traits continue.

Which of the following defines a genome?
A) representation of a complete set of a cell's polypeptides

B) the complete set of an organism's polypeptides
C) the complete set of a species' polypeptides
D) a karyotype
E) the complete set of an organism's genes

Which is the smallest unit containing the entire human genome?

A) one human somatic cell
B) one human chromosome
C) all of the DNA of one human

D) the entire human population

E) one human gene

If an organism is diploid and a certain gene found in the organism has 18 known alleles (variants), then any given organism of that species can/must have which of the following?

A) at most, 2 alleles for that gene
B) up to 18 chromosomes with that gene

C) up to 18 genes for that trait
D) a haploid number of 9 chromosomes
E) up to, but not more than, 18 different traits

Which of the following is a true statement about sexual vs. asexual reproduction?
A) Asexual reproduction, but not sexual reproduction, is characteristic of plants and fungi.

B) In sexual reproduction, individuals transmit 50% of their genes to each of their offspring.

C) In asexual reproduction, offspring are produced by fertilization without meiosis.
D) Sexual reproduction requires that parents be diploid.
E) Asexual reproduction produces only haploid offspring.

At which stage of mitosis are chromosomes usually photographed in the preparation of a karyotype?
A) prophase
B) metaphase

C) anaphase
D) telophase
E) interphase

Which of the following is true of a species that has a chromosome number of 2n = 16? A) The species is diploid with 32 chromosomes per cell.
B) The species has 16 sets of chromosomes per cell.
C) Each cell has eight homologous pairs.

D) During the S phase of the cell cycle there will be 32 separate chromosomes.

E) A gamete from this species has four chromosomes.

Eukaryotic sexual life cycles show tremendous variation. Of the following elements, which do all sexual life cycles have in common?

1. Alternation of generations
2. Meiosis
3. Fertilization
4. Gametes
5. Spores

A) I, IV, and V
B) I, II, and IV
C) II, III, and IV
D) II, IV, and V
E) I, II, III, IV, and V

Which of these statements is false?
A) In humans, each of the 22 maternal autosomes has a homologous paternal chromosome.

B) In humans, the 23rd pair, the sex chromosomes, determines whether the person is female (XX) or male (XY).
C) Single, haploid (n) sets of chromosomes in ovum and sperm unite during fertilization, forming a diploid (2n), single-celled zygote.
D) At sexual maturity, ovaries and testes produce diploid gametes by meiosis.
E) Sexual life cycles differ with respect to the relative timing of meiosis and fertilization.

Referring to a plant's sexual life cycle, which of the following terms describes the process that leads directly to the formation of gametes?
A) sporophyte meiosis
B) gametophyte mitosis

C) gametophyte meiosis
D) sporophyte mitosis
E) alternation of generations

Which of the following is an example of alternation of generations?
A) A grandparent and grandchild each have dark hair, but the parent has blond hair.
B) A diploid plant (sporophyte) produces, by meiosis, a spore that gives rise to a multicellular, haploid pollen grain (gametophyte).
C) A diploid animal produces gametes by meiosis, and the gametes undergo fertilization to produce a diploid zygote.
D) A haploid mushroom produces gametes by mitosis, and the gametes undergo fertilization, which is immediately followed by meiosis.
E) A diploid cell divides by mitosis to produce two diploid daughter cells, which then fuse to produce a tetraploid cell.

The human X and Y chromosomes
A) are both present in every somatic cell of males and females alike.
B) are about the same size and have approximately the same number of genes.

C) are almost entirely homologous, despite their different names.
D) include genes that determine an individual's sex.
E) are called autosomes.

Which of these is a karyotype?
A) a natural cellular arrangement of chromosomes in the nucleus

B) a display of all the cell types in an organism
C) organized images of a cell’s chromosomes
D) the appearance of an organism
E) a display of a cell’s mitotic stages

Mitosis is commonly found in all of the following except

A) a haploid animal cell.
B) a diploid animal cell.
C) a haploid plant cell.

D) a diploid plant cell.

Which of these is a way that the sexual life cycle increases genetic variation in a species?

A) by allowing crossing over
B) by allowing an increase in cell number
C) by increasing gene stability

D) by conserving chromosomal gene order

E) by decreasing mutation frequency

A given organism has 46 chromosomes in its karyotype. We can therefore conclude which of the following?
A) It must be human.
B) It must be a primate.

C) It must be an animal.
D) It must be sexually reproducing.
E) Its gametes must have 23 chromosomes.

A triploid cell contains three sets of chromosomes. If a cell of a usually diploid species with 42 chromosomes per cell is triploid, this cell would be expected to have which of the following?

A) 63 chromosomes in 31 1/2 pairs
B) 63 chromosomes in 21 sets of 3

C) 63 chromosomes, each with three chromatids
D) 21 chromosome pairs and 21 unique chromosomes

Which of the following best describes a karyotype?
A) a pictorial representation of all the genes for a species
B) a display of each of the chromosomes of a single cell
C) the combination of all the maternal and paternal chromosomes of a species

D) the collection of all the chromosomes in an individual organism
E) a photograph of all the cells with missing or extra chromosomes

Which of the following can utilize both mitosis and meiosis in the correct circumstances?

A) a haploid animal cell
B) a diploid cell from a plant stem
C) any diploid animal cell

D) a plantlike protist
E) an archaebacterium

The somatic cells of a privet shrub each contain 46 chromosomes. To be as different as they are from human cells, which have the same number of chromosomes, which of the following must be true?
A) Privet cells cannot reproduce sexually.

B) Privet sex cells have chromosomes that can synapse with human chromosomes in the laboratory.
C) Genes of privet chromosomes are significantly different than those in humans.
D) Privet shrubs must be metabolically more like animals than like other shrubs.

E) Genes on a particular privet chromosome, such as the X, must be on a different human chromosome, such as number 18.

In a human karyotype, chromosomes are arranged in 23 pairs. If we choose one of these pairs, such as pair 14, which of the following do the two chromosomes of the pair have in common?
A) length and position of the centromere only

B) length, centromere position, and staining pattern only
C) length, centromere position, staining pattern, and traits coded for by their genes

D) length, centromere position, staining pattern, and DNA sequences
E) They have nothing in common except they are X-shaped.

To view and analyze human chromosomes in a dividing cell, which of the following is (are) required?
A) a scanning electron microscope
B) radioactive staining

C) fluorescent staining and a transmission electron microscope

D) DNA staining and a light microscope
E) a stain particular to human cells

The karyotype of one species of primate has 48 chromosomes. In a particular female, cell division goes awry and she produces one of her eggs with an extra chromosome (25). The most probable source of this error would be a mistake in which of the following?
A) mitosis in her ovary

B) metaphase I of one meiotic event

C) telophase II of one meiotic event

D) telophase I of one meiotic event

E) either anaphase I or II

If a cell has completed the first meiotic division and is just beginning meiosis II, which of the following is an appropriate description of its contents?
A) It has half the amount of DNA as the cell that began meiosis.
B) It has the same number of chromosomes but each of them has different alleles than another cell from the same meiosis.

C) It has half the chromosomes but twice the DNA of the originating cell.
D) It has one-fourth the DNA and one-half the chromosomes as the originating cell.

E) It is identical in content to another cell from the same meiosis.

Which of the following might result in a human zygote with 45 chromosomes?
A) an error in either egg or sperm meiotic anaphase
B) failure of the egg nucleus to be fertilized by the sperm
C) fertilization of a 23 chromosome human egg by a 22 chromosome sperm of a closely related species

D) an error in the alignment of chromosomes on the metaphase plate

E) lack of chiasmata in prophase I

After telophase I of meiosis, the chromosomal makeup of each daughter cell is A) diploid, and the chromosomes are each composed of a single chromatid.
B) diploid, and the chromosomes are each composed of two chromatids.
C) haploid, and the chromosomes are each composed of a single chromatid.

D) haploid, and the chromosomes are each composed of two chromatids.

E) tetraploid, and the chromosomes are each composed of two chromatids.

How do cells at the completion of meiosis compare with cells that have replicated their DNA and are just about to begin meiosis?
A) They have twice the amount of cytoplasm and half the amount of DNA.
B) They have half the number of chromosomes and half the amount of DNA.

C) They have the same number of chromosomes and half the amount of DNA.
D) They have half the number of chromosomes and one-fourth the amount of DNA.

E) They have half the amount of cytoplasm and twice the amount of DNA.

When does the synaptonemal complex disappear?

A) late prophase of meiosis I
B) during fertilization or fusion of gametes
C) early anaphase of meiosis I

D) mid-prophase of meiosis II

E) late metaphase of meiosis II

Which of the following happens at the conclusion of meiosis I?

A) Homologous chromosomes of a pair are separated from each other.

B) The chromosome number per cell is conserved.
C) Sister chromatids are separated.
D) Four daughter cells are formed.
E) Cohesins are cleaved at the centromeres.

Chromatids are separated from each other.
A) The statement is true for mitosis only.
B) The statement is true for meiosis I only.
C) The statement is true for meiosis II only.
D) The statement is true for mitosis and meiosis I.

E) The statement is true for mitosis and meiosis II.

Which of the following occurs in meiosis but not in mitosis?

A) chromosome replication
B) synapsis of chromosomes
C) production of daughter cells

D) alignment of chromosomes at the equator E) condensation of chromatin

Whether during mitosis or meiosis, sister chromatids are held together by proteins referred to as cohesins. Such molecules must have which of the following properties?
A) They must persist throughout the cell cycle.
B) They must be removed before meiosis can begin.

C) They must be removed before sister chromatids or homologous chromosomes can separate.

D) They must reattach to chromosomes during G1.
E) They must be intact for nuclear envelope re-formation.

Experiments with cohesins have found that
A) cohesins are protected from destruction throughout meiosis I and II.
B) cohesins are cleaved from chromosomes at the centromere before anaphase I.

C) cohesins are protected from cleavage at the centromere during meiosis I.
D) a protein cleaves cohesins before metaphase I.
E) a protein that cleaves cohesins would cause cellular death.

A pair of homologous chromosomes includes which of the following sets of DNA strands? A) two single-stranded chromosomes that have synapsed
B) two sister chromatids that have synapsed
C) four sister chromatids

D) four unique chromosomes

E) eight sister chromatids

When we see chiasmata under a microscope, that lets us know which of the following has occurred?
A) asexual reproduction
B) meiosis II

C) anaphase II
D) prophase I
E) separation of homologs

To visualize and identify meiotic cells at metaphase with a microscope, what would you look for?
A) sister chromatids of a replicated chromosome grouped at the poles
B) individual chromosomes all at the cell's center

C) an uninterrupted spindle array
D) the synaptonemal complex
E) pairs of homologous chromosomes all aligned at the cell's center

For the following questions, match the key event of meiosis with the stages listed below.

1. Prophase I
2. Metaphase I
3. Anaphase I
4. Telophase I

V. Prophase II VI. Metaphase II VII. Anaphase II VIII. Telophase II

Centromeres of sister chromatids disjoin and chromatids separate.

A) II
B) III
C) IV

D) V
E) VII

For the following questions, match the key event of meiosis with the stages listed below.

1. Prophase I
2. Metaphase I
3. Anaphase I
4. Telophase I

V. Prophase II VI. Metaphase II VII. Anaphase II VIII. Telophase II

Homologous chromosomes are aligned at the equator of the spindle.

A) I
B) II
C) IV

D) VI
E) VIII

For the following questions, match the key event of meiosis with the stages listed below.

1. Prophase I
2. Metaphase I
3. Anaphase I
4. Telophase I

V. Prophase II VI. Metaphase II VII. Anaphase II VIII. Telophase II

Synaptonemal complexes form or are still present.

A) I only
B) I and IV only
C) I and VIII only

D) II and VI only
E) I, II, III, and IV only

The following question refers to the essential steps in meiosis described below.

1. Formation of four new nuclei, each with half the chromosomes present in the parental nucleus 2. Alignment of homologous chromosomes at the metaphase plate
3. Separation of sister chromatids
4. Separation of the homologs; no uncoupling of the centromere

5. Synapsis; chromosomes moving to the middle of the cell in pairs

Which of the steps take(s) place in both mitosis and meiosis?

A) 2
B) 3
C) 5

D) 2 and 3 only
E) 2, 3, and 5

For a species with a haploid number of 23 chromosomes, how many different combinations of maternal and paternal chromosomes are possible for the gametes?
A) 23
B) 46

C) 460
D) 920
E) about 8 million

Independent assortment of chromosomes is a result of
A) the random and independent way in which each pair of homologous chromosomes lines up at the metaphase plate during meiosis I.
B) the random nature of the fertilization of ova by sperm.
C) the random distribution of the sister chromatids to the two daughter cells during anaphase II.

D) the relatively small degree of homology shared by the X and Y chromosomes.
E) the random and independent way in which each pair of homologous chromosomes lines up at the metaphase plate during meiosis I, the random nature of the fertilization of ova by sperm, the random distribution of the sister chromatids to the two daughter cells during anaphase II, and the relatively small degree of homology shared by the X and Y chromosomes.

Independent assortment of chromosomes occurs.

A) The statement is true for mitosis only.
B) The statement is true for meiosis I only.
C) The statement is true for meiosis II only.

D) The statement is true for mitosis and meiosis I.

E) The statement is true for mitosis and meiosis II.

Which of the following best describes the frequency of crossing over in mammals?

A) ~50 per chromosome pair
B) ~2 per meiotic cell
C) at least 1-2 per chromosome pair

D) ~1 per pair of sister chromatids
E) a very rare event among hundreds of cells

When homologous chromosomes cross over, what occurs?
A) Two chromatids get tangled, resulting in one re-sequencing its DNA.
B) Two sister chromatids exchange identical pieces of DNA.
C) Specific proteins break the two strands of nonsister chromatids and re-join them.
D) Each of the four DNA strands of a homologous pair is broken, and the pieces are mixed.

E) Maternal alleles are "corrected" to be like paternal alleles, and vice versa.

In part III of Figure 10.1, the progression of events corresponds to which of the following series?
A) zygote, mitosis, gametophyte, mitosis, fertilization, zygote, mitosis
B) sporophyte, meiosis, spore, mitosis, gametophyte, mitosis, gametes, fertilization

C) fertilization, mitosis, multicellular haploid, mitosis, spores, sporophyte

D) gametophyte, meiosis, zygote, spores, sporophyte, zygote
E) meiosis, fertilization, zygote, mitosis, adult, meiosis

In a life cycle such as that shown in part III of Figure 10.1, if the zygote's chromosome number is 10, which of the following will be true?
A) The sporophyte's chromosome number per cell is 10 and the gametophyte's is 5.
B) The sporophyte's chromosome number per cell is 5 and the gametophyte's is 10.

C) The sporophyte and gametophyte each have 10 chromosomes per cell.

D) The sporophyte and gametophyte each have 5 chromosomes per cell.

E) The sporophyte and gametophyte each have 20 chromosomes per cell.

A certain female's number 12 chromosomes both have the blue gene and number 19 chromosomes both have the long gene. As cells in her ovaries undergo meiosis, her resulting eggs (ova) may have which of the following?
A) either two number 12 chromosomes with blue genes or two with orange genes

B) either two number 19 chromosomes with long genes or two with short genes
C) either one blue or one orange gene in addition to either one long or one short gene D) one chromosome 12 with one blue gene and one chromosome 19 with one long gene

If a female of this species has one chromosome 12 with a blue gene and another chromosome 12 with an orange gene, and has both number 19 chromosomes with short genes, she will produce which of the following egg types?
A) only blue short gene eggs

B) only orange short gene eggs
C) one-half blue short and one-half orange short gene eggs
D) three-fourths blue long and one-fourth orange short gene eggs

E) three-fourths blue short and one-fourth orange short gene eggs

A female with a paternal set of one orange and one long gene chromosome and a maternal set comprised of one blue and one short gene chromosome is expected to produce which of the following types of eggs after meiosis?
A) All eggs will have maternal types of gene combinations.

B) All eggs will have paternal types of gene combinations.
C) Half the eggs will have maternal and half will have paternal combinations.
D) Each egg has a one-fourth chance of having either blue long, blue short, orange long, or orange short combinations.
E) Each egg has a three-fourths chance of having blue long, one-fourth blue short, three-fourths orange long, or one-fourth orange short combinations.

Because the rotifers develop from eggs, but asexually, what can you predict?

A) The eggs and the zygotes are all haploid.
B) The animals are all hermaphrodites.
C) Although asexual, both males and females are found in nature.

D) All males can produce eggs.

E) No males can be found.

How is natural selection related to sexual reproduction as opposed to asexual reproduction?

A) Sexual reproduction results in many new gene combinations, some of which will lead to differential reproduction.
B) Sexual reproduction results in the most appropriate and healthiest balance of two sexes in a population.

C) Sexual reproduction results in the greatest number of new mutations.
D) Sexual reproduction allows the greatest number of offspring to be produced.

E) Sexual reproduction utilizes far less energy than asexual reproduction.

A human cell containing 22 autosomes and a Y chromosome is

A) a sperm.
B) an egg.
C) a zygote.

D) a somatic cell of a male.
E) a somatic cell of a female.

Homologous chromosomes move toward opposite poles of a dividing cell during

A) mitosis.
B) meiosis I.
C) meiosis II.

D) fertilization.
E) binary fission

If the DNA content of a diploid cell in the G1 phase of the cell cycle is x, then the DNA content of the same cell at metaphase of meiosis I would be

A) 0.25x.

B) 0.5x.

C) x.
D) 2x.

E) 4x.

If we continued to follow the cell lineage from question 3, then the DNA content of a single cell at metaphase of meiosis II would be
A) 0.25x.
B) 0.5x.

C) x.
D) 2x.
E) 4x

How many different combinations of maternal and paternal chromosomes can be packaged in gametes made by an organism with a diploid number of 8 (2n = 8)?
A) 2
B) 4

C) 8
D) 16
E) 32

What do we mean when we use the terms monohybrid cross and dihybrid cross?
A) A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents.

B) A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny.
C) A dihybrid cross involves organisms that are heterozygous for two characters, and a monohybrid cross involves only one.
D) A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations.
E) A monohybrid cross results in a 9:3:3:1 ratio, whereas a dihybrid cross gives a 3:1 ratio.

What was the most significant conclusion that Gregor Mendel drew from his experiments with pea plants?
A) There is considerable genetic variation in garden peas.
B) Traits are inherited in discrete units, and are not the results of "blending."

C) Recessive genes occur more frequently in the F1 generation than do dominant ones.

D) Genes are composed of DNA.

E) An organism that is homozygous for many recessive traits is at a disadvantage.

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?
A) 4
B) 8

C) 16
D) 32
E) 64

The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason?
A) segregation of maternal and paternal alleles
B) recurrent mutations forming new alleles

C) crossing over during prophase I
D) different possible assortment of chromosomes into gametes

E) the tendency for dominant alleles to segregate together

Why did Mendel continue some of his experiments to the F2 or F3 generation?

A) to obtain a larger number of offspring on which to base statistics

B) to observe whether or not a recessive trait would reappear

C) to observe whether or not the dominant trait would reappear

D) to distinguish which alleles were segregating
E) to be able to describe the frequency of recombination

Which of the following differentiates between independent assortment and segregation?

A) The law of independent assortment requires describing two or more genes relative to one another.
B) The law of segregation requires describing two or more genes relative to one another.

C) The law of segregation requires having two or more generations to describe.
D) The law of independent assortment is accounted for by observations of prophase I. E) The law of segregation is accounted for by anaphase of mitosis.

Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular trait. What does this suggest?
A) that the parents were true-breeding for contrasting traits
B) that the trait shows incomplete dominance

C) that a blending of traits has occurred
D) that the parents were both heterozygous for a single trait
E) that each offspring has the same alleles for each of two traits

A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism?
A) tt

B) Hh
C) HhTt
D) T
E) HT

When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype?
A) 0%

B) 25%
C) 50%
D) 75%
E) 100%

Mendel accounted for the observation that traits that had disappeared in the F1 generation reappeared in the F2 generation by proposing that
A) new mutations were frequently generated in the F2 progeny, "reinventing" traits that had been lost in the F1.
B) the mechanism controlling the appearance of traits was different between the F1 and the F2 plants.
C) traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1.
D) the traits were lost in the F1 due to dominance of the parental traits.
E) members of the F1 generation had only one allele for each trait, but members of the F2 had two alleles for each trait.

The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following?
A) None of the traits obeyed the law of segregation.
B) The diploid number of chromosomes in the pea plants was 7.

C) All of the genes controlling the traits were located on the same chromosome.
D) All of the genes controlling the traits behaved as if they were on different chromosomes.

E) The formation of gametes in plants occurs by mitosis only.

Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division?
A) prophase I of meiosis
B) anaphase II of meiosis

C) metaphase I of meiosis

D) anaphase I of meiosis

E) anaphase of mitosis

Mendel's second law of independent assortment has its basis in which of the following events of meiosis I?
A) synapsis of homologous chromosomes
B) crossing over

C) alignment of tetrads at the equator

D) separation of homologs at anaphase

E) separation of cells at telophase

Why did the F1 offspring of Mendel's classic pea cross always look like one of the two parental varieties?

A) No genes interacted to produce the parental phenotype.

B) Each allele affected phenotypic expression.
C) The traits blended together during fertilization.
D) One allele was dominant.

E) Phenotype was not dependent on genotype.

Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of crosses BbTt × BBtt will be expected to have black fur and long tails?
A) 1/16

B) 3/16
C) 3/8
D) 1/2
E) 9/16

In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short?
A) 1
B) 1/2

C) 1/4
D) 1/6
E) 0

In the cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC?

A) 1/4
B) 1/8
C) 1/16

D) 1/32
E) 1/64

Given the parents AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent?
A) 1/4

B) 1/8
C) 3/4
D) 3/8
E) 1

Which of the following is the best statement of the use of the addition rule of probability?
A) the probability that two or more independent events will both occur
B) the probability that two or more independent events will both occur in the offspring of one set of parents
C) the probability that either one of two independent events will occur
D) the probability of producing two or more heterozygous offspring
E) the likelihood that a trait is due to two or more meiotic events

Which of the following calculations require that you utilize the addition rule?
A) Calculate the probability of black offspring from the cross AaBb × AaBb, when B is the symbol for black.
B) Calculate the probability of children with both cystic fibrosis and polydactyly when parents are each heterozygous for both genes.
C) Calculate the probability of each of four children having cystic fibrosis if the parents are both heterozygous.
D) Calculate the probability of a child having either sickle-cell anemia or cystic fibrosis if parents are each heterozygous for both.
E) Calculate the probability of purple flower color in a plot of 50 plants seeded from a self- fertilizing heterozygous parent plant.

Marfan syndrome in humans is caused by an abnormality of the connective tissue protein fibrillin. Patients are usually very tall and thin, with long spindly fingers, curvature of the spine, sometimes weakened arterial walls, and sometimes ocular problems, such as lens dislocation. Which of the following would you conclude about Marfan syndrome from this information?

A) It is recessive.
B) It is dominant.
C) It has a late age of onset (> 60).

D) It is pleiotropic.
E) It is epistatic.

In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white?
A) red × white

B) roan × roan
C) white × roan
D) red × roan
E) The answer cannot be determined from the information provided.

Which of the following describes the ability of a single gene to have multiple phenotypic effects?
A) incomplete dominance
B) multiple alleles

C) pleiotropy
D) epistasis

Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this?
A) incomplete dominance
B) multiple alleles
C) pleiotropy
D) epistasis
E) codominance

Which of the following is an example of polygenic inheritance?

A) pink flowers in snapdragons
B) the ABO blood group in humans
C) Huntington's disease in humans

D) white and purple flower color in peas

E) skin pigmentation in humans

Hydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by which of the following?
A) the knowledge that multiple alleles are involved

B) the allele for blue hydrangea being completely dominant

C) the alleles being codominant
D) the fact that a mutation has occurred
E) environmental factors such as soil pH

Which of the following provides an example of epistasis?
A) Recessive genotypes for each of two genes (aabb) result in an albino corn snake.
B) The allele b17 produces a dominant phenotype, although b1 through b16 do not.
C) In rabbits and many other mammals, one genotype (ee) prevents any fur color from developing.
D) In Drosophila (fruit flies), white eyes can be due to an X-linked gene or to a combination of other genes.
E) In cacti, there are several genes for the type of spines.

A scientist discovers a DNA-based test for one allele of a particular gene. This and only this allele, if homozygous, produces an effect that results in death at or about the time of birth. Of the following, which is the best use of this discovery?
A) Screen all newborns of an at-risk population.

B) Design a test for identifying heterozygous carriers of the allele.

C) Introduce a normal allele into deficient newborns.
D) Follow the segregation of the allele during meiosis.
E) Test school-age children for the disorder.

The frequency of heterozygosity for the sickle-cell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following?
A) Mendel's law of independent assortment
B) Mendel's law of segregation
C) Darwin's explanation of natural selection
D) Darwin's observations of competition
E) the malarial parasite changing the allele

One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. If a young child is the first in her family to be diagnosed, which of the following is the best explanation?
A) The mother carries the gene but does not express it at all.

B) One of the parents has very mild expression of the gene.

C) The condition skipped a generation in the family.
D) The child has a different allele of the gene than the parents.

In each generation of this family after generation I, the age at diagnosis is significantly lower than would be found in nonfamilial (sporadic) cases of this cancer (~63 years). What is the most likely reason?
A) Members of this family know to be checked for colon cancer early in life.

B) Hereditary (or familial) cases of this cancer typically occur at earlier ages than do nonfamilial forms.
C) This is pure chance; it would not be expected if you were to look at a different family.
D) This cancer requires mutations in more than this one gene.

E) Affected members of this family are born with colon cancer, and it can be detected whenever they are first tested.

The affected woman in generation IV is thinking about her future and asks her oncologist (cancer specialist) whether she can know whether any or all of her children will have a high risk of the same cancer. The doctor would be expected to advise which of the following?
I. genetic counseling
II. prenatal diagnosis when/if she becomes pregnant
III. testing to see whether she has the allele
IV.testing to see whether her future spouse or partner has the allele
A) I only
B) II only
C) I, II, and IV only
D) I, II, and III only
E) III and IV only

Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. The genes for flower color and location assort independently.

If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers?

A) 65
B) 190
C) 250
D) 565
E) 750

Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. The genes for flower color and location assort independently.

Among the F2 offspring, what is the probability of plants with white axial flowers? A) 9/16
B) 1/16
C) 3/16

D) 1/8
E) 1/4

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.

These results indicate which of the following? A) Brown is dominant to black.

B) Black is dominant to brown and to yellow.
C) Yellow is dominant to black.

D) There is incomplete dominance.

E) Epistasis is involved.

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.

How many genes must be responsible for these coat colors in Labrador retrievers?

A) 1
B) 2
C) 3

D) 4
E) 5

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.

In one type cross of black × black, the results were as follows:

9/16 black
4/16 yellow
3/16 brown

The genotype eebb must result in which of the following?

A) black
B) brown
C) yellow

D) a lethal result
E) white

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white- flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait.

If true-breeding red long radishes are crossed with true-breeding white oval radishes, the F1 will be expected to be which of the following?

A) red and long
B) red and oval

C) white and long

D) purple and long

E) purple and oval

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white- flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait.

In the F2 generation of the above cross, which of the following phenotypic ratios would be expected?

A) 9:3:3:1
B) 9:4:3
C) 1:1:1:1
D) 1:1:1:1:1:1
E) 6:3:3:2:1:1

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white- flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait.

The flower color trait in radishes is an example of which of the following?

A) a multiple allelic system
B) sex linkage
C) codominance

D) incomplete dominance

E) epistasis

Drosophila (fruit flies) usually have long wings (+), but mutations in two different genes can

result in bent wings (bt) or vestigial wings (vg).

If a homozygous bent wing fly is mated with a homozygous vestigial wing fly, which of the following offspring would you expect?

A) all +bt +vg heterozygotes
B) 1/2 bent and 1/2 vestigial flies

C) all homozygous + flies
D) 3/4 bent to 1/4 vestigial ratio
E) 1/2 bent and vestigial to 1/2 normal

Drosophila (fruit flies) usually have long wings (+), but mutations in two different genes can

result in bent wings (bt) or vestigial wings (vg).

If flies that are heterozygous for both the bent wing gene and the vestigial wing gene are mated, what is the probability of offspring with bent wings only?
A) 1/8
B) 3/8

C) 1/4
D) 9/16
E) 3/16

A dwarf, red snapdragon is crossed with a plant homozygous for tallness and white flowers. What are the genotype and phenotype of the F1 individuals?
A) ttRr–dwarf and pink
B) ttrr–dwarf and white

C) TtRr–tall and red

D) TtRr–tall and pink

E) TTRR–tall and red

If snapdragons are heterozygous for height as well as for flower color, a mating between them will result in what ratio?
A) 9:3:3:1
B) 6:3:3:2:1:1

C) 1:2:1
D) 27:9:9:9:3:3:3:1
E) 9:4:3

How many different types of gametes would be possible in this system?

A) 1
B) 2
C) 4

D) 8
E) 16

One fish of this type has alleles 1 and 3 (S1S3) and its mate has alleles 2 and 4 (S2S4). If each allele confers a unit of color darkness such that S1 has one unit, S2 has two units, and so on,

then what proportion of their offspring would be expected to have five units of color?

A) 1/4
B) 1/5
C) 1/8

D) 1/2
E) 0

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.

The relationship between genes S and N is an example of

A) incomplete dominance.
B) epistasis.
C) complete dominance.

D) pleiotropy.
E) codominance.

A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce

A) all sharp-spined progeny.
B) 50% sharp-spined, 50% dull-spined progeny.
C) 25% sharp-spined, 50% dull-spined, 25% spineless progeny.

D) all spineless progeny.
E) It is impossible to determine the phenotypes of the progeny.

If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios?

A) 3 sharp-spined:1 spineless
B) 1 sharp-spined:2 dull-spined:1 spineless C) 1 sharp-spined:1 dull-spined:1 spineless D) 1 sharp-spined:1 dull-spined
E) 9 sharp-spined:3 dull-spined:4 spineless

A blue budgie is crossed with a white budgie. Which of the following results is not possible? A) green offspring only
B) yellow offspring only
C) blue offspring only

D) green and yellow offspring

E) a 9:3:3:1 ratio

Two blue budgies were crossed. Over the years, they produced 22 offspring, 5 of which were white. What are the most likely genotypes for the two blue budgies?
A) yyBB and yyBB
B) yyBB and yyBb

C) yyBb and yyBb
D) yyBB and yybb
E) yyBb and yybb

Which of the following is a possible partial genotype for the son?

A) IBIB
B) IBIA
C) ii

D) IBi

E) IAIA

Which of the following is a possible genotype for the mother?

A) IAIA

B) IBIB

C) ii
D) IAi

E) IAIB

Which of the following is a possible phenotype for the father?

A) A negative
B) O negative
C) B positive

D) AB negative
E) impossible to determine

If both children are of blood type M, which of the following is possible?

A) Each parent is either M or MN.
B) Each parent must be type M.
C) Both children are heterozygous for this gene.

D) Neither parent can have the N allele.
E) The MN blood group is recessive to the ABO blood group.

One species of a small birdlike animal has an extremely variable tail length, an example of polygenic inheritance. Geneticists have come to realize that there are eight separate genes for tail length per haploid genome, with each gene having two alleles. One allele for each gene (a1, b1, and so on) increases the length by 1 cm, whereas the other allele (a2, b2, and so on) increases it by 0.5 cm. One bird was analyzed and found to have the following genotype: a1a1b2b2c1c2d1d2e2e2f1f2g1g1h1h2

What is the length of its tail?

A) 6 cm
B) 8 cm
C) 12 cm

D) 24 cm
E) 36 cm

One species of green plant, with frondlike leaves, a spine-coated stem, and purple cup- shaped flowers, is found to be self-pollinating. Which of the following is true of this species? A) The species must be haploid.
B) Its reproduction is asexual.

C) All members of the species have the same genotype.

D) Some of the seeds would have true-breeding traits.

E) All of its dominant traits are most frequent.

If the environmental parameters, such as temperature, humidity, atmosphere, sunlight, and so on, are mostly Earthlike, which of the following do you expect of its types of leaves, stems, and flowers?
A) The genes for them would have originated on Earth.

B) Genes for these traits would have a common ancestor with those from Earth.
C) Such plants could be safely eaten by humans.
D) Genotypes for these traits would be identical to those of Earth plants with the same traits.

E) Phenotypes would be selected for or against by these environmental factors.

When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?
A) The gene involved is on the Y chromosome.

B) The gene involved is on the X chromosome.
C) The gene involved is on an autosome, but only in males.

D) Other male-specific factors influence eye color in flies.

E) Other female-specific factors influence eye color in flies.

Which of the following is the meaning of the chromosome theory of inheritance as expressed in the early 20th century?
A) Individuals inherit particular chromosomes attached to genes.
B) Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis.

C) Homologous chromosomes give rise to some genes and crossover chromosomes to other genes.

D) No more than a single pair of chromosomes can be found in a healthy normal cell.

E) Natural selection acts on certain chromosome arrays rather than on genes.

Males are more often affected by sex-linked traits than females because
A) male hormones such as testerone often alter the affects of mutations on the X chromosome.

B) female hormones such as estrogen often compensate for the effects of mutations on the X chromosome.
C) X chromosomes in males generally have more mutations than X chromosomes in females.

D) males are hemizygous for the X chromosome.
E) mutations on the Y chromosome often worsen the effects of X-linked mutations.

SRY is best described in which of the following ways?
A) a gene present on the X chromosome that triggers female development
B) an autosomal gene that is required for the expression of genes on the Y chromosome
C) a gene region present on the Y chromosome that triggers male development
D) an autosomal gene that is required for the expression of genes on the X chromosome
E) a gene required for development, and males or females lacking the gene do not survive past early childhood

In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male?
A) tortoiseshell females; tortoiseshell males

B) black females; orange males
C) orange females; orange males
D) tortoiseshell females; black males

E) orange females; black males

Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents?

A) XnXn and XnY

B) XnXn and XNY

C) XNXN and XnY

D) XNXN and XNY

E) XNXn and XNY

Cinnabar eyes is a sex-linked recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a wild-type male, what percentage of the F1 males will have cinnabar eyes?

A) 0%
B) 25%

C) 50%
D) 75%
E) 100%

Normally, only female cats have the tortoiseshell phenotype because
A) the males die during embryonic development.
B) a male inherits only one allele of the X-linked gene controlling hair color.
C) the Y chromosome has a gene blocking orange coloration.
D) only males can have Barr bodies.
E) multiple crossovers on the Y chromosome prevent orange pigment production.

Sex determination in mammals is due to the SRY region of the Y chromosome. An abnormality of this region could allow which of the following to have a male phenotype? A) Turner syndrome, 45, X
B) translocation of SRY to an autosome of a 46, XX individual
C) a person with an extra X chromosome
D) a person with one normal and one shortened (deleted) X
E) Down syndrome, 46, XX

In humans, clear gender differentiation occurs not at fertilization, but after the second month of gestation. What is the first event of this differentiation?
A) formation of testosterone in male embryos
B) formation of estrogens in female embryos

C) anatomical differentiation of a penis in male embryos
D) activation of SRY in male embryos and masculinization of the gonads E) activation of SRY in females and feminization of the gonads

Duchenne muscular dystrophy is a serious condition caused by a recessive allele of a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their 20s. How likely is it for a woman to have this condition?

A) Women can never have this condition.
B) One-half of the daughters of an affected man would have this condition.
C) One-fourth of the daughters of an affected father and a carrier mother could have this condition.
D) Very rarely: it is rare that an affected male would mate with a carrier female.
E) Only if a woman is XXX could she have this condition.

All female mammals have one active X chromosome per cell instead of two. What causes this?
A) activation of the XIST gene on the X chromosome that will become the Barr body
B) activation of the BARR gene on one X chromosome, which then becomes inactive

C) crossing over between the XIST gene on one X chromosome and a related gene on an autosome
D) inactivation of the XIST gene on the X chromosome derived from the male parent
E) attachment of methyl (CH3) groups to the X chromosome that will remain active

Which of the following statements is true of linkage?
A) The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them.
B) The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%.
C) All of the traits that Mendel studied–seed color, pod shape, flower color, and others–are due to genes linked on the same chromosome.
D) Linked genes are found on different chromosomes.
E) Crossing over occurs during prophase II of meiosis.

How would one explain a testcross involving F1 dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced?

A) The two genes are closely linked on the same chromosome.

B) The two genes are linked but on different chromosomes.
C) Recombination did not occur in the cell during meiosis.
D) The testcross was improperly performed.

E) Both of the characters are controlled by more than one gene.

What does a frequency of recombination of 50% indicate?
A) The two genes are likely to be located on different chromosomes.
B) All of the offspring have combinations of traits that match one of the two parents.

C) The genes are located on sex chromosomes.
D) Abnormal meiosis has occurred.
E) Independent assortment is hindered.

Three genes (A, B, and C) at three loci are being mapped in a particular species. Each gene has two alleles, one of which results in a phenotype that is markedly different from the wild type. The unusual allele of gene A is inherited with the unusual allele of gene B or C about 50% of the time. However, the unusual alleles of genes B and C are inherited together 14.4% of the time. Which of the following describes what is happening?
A) The three genes are showing independent assortment.
B) The three genes are linked.
C) Gene A is linked but genes B and C are not.
D) Gene A is assorting independently of genes B and C, which are linked.
E) Gene A is located 14.4 map units from genes B and C.

What is one map unit equivalent to?
A) the physical distance between two linked genes
B) 1% frequency of recombination between two genes
C) 1 nanometer of distance between two genes
D) the distance between a pair of homologous chromosomes
E) the recombination frequency between two genes assorting independently

Recombination between linked genes comes about for what reason?
A) Mutation on one homolog is different from that on the other homolog.
B) Independent assortment sometimes fails because Mendel had not calculated appropriately.

C) When genes are linked they always "travel" together at anaphase.
D) Crossovers between these genes result in chromosomal exchange.
E) Nonrecombinant chromosomes break and then re-join with one another.

Why does recombination between linked genes continue to occur?

A) Recombination is a requirement for independent assortment.
B) Recombination must occur or genes will not assort independently.

C) New allele combinations are acted upon by natural selection.

D) The forces on the cell during meiosis II always result in recombination.

E) Without recombination there would be an insufficient number of gametes.

Map units on a linkage map cannot be relied upon to calculate physical distances on a chromosome for which of the following reasons?
A) The frequency of crossing over varies along the length of the chromosome.
B) The relationship between recombination frequency and map units is different in every individual.

C) Physical distances between genes change during the course of the cell cycle.

D) The gene order on the chromosomes is slightly different in every individual.

E) Linkage map distances are identical between males and females.

What is the reason that closely linked genes are typically inherited together?

A) The likelihood of a crossover event between these two genes is low.
B) The number of genes in a cell is greater than the number of chromosomes.

C) Chromosomes are unbreakable.

D) Alleles are paired together during meiosis.
E) Genes align that way during metaphase I of meiosis.

Sturtevant provided genetic evidence for the existence of four pairs of chromosomes in Drosophila in which of these ways?
A) There are four major functional classes of genes in Drosophila.
B) Drosophila genes cluster into four distinct groups of linked genes.

C) The overall number of genes in Drosophila is a multiple of four.

D) The entire Drosophila genome has approximately 400 map units.

E) Drosophila genes have, on average, four different alleles.

If cell X enters meiosis, and nondisjunction of one chromosome occurs in one of its daughter cells during meiosis II, what will be the result at the completion of meiosis?
A) All the gametes descended from cell X will be diploid.
B) Half of the gametes descended from cell X will be n + 1, and half will be n - 1.

C) One-fourth of the gametes descended from cell X will be n + 1, 1/4 will be n - 1, and 1/2 will be n.
D) There will be three extra gametes.
E) Two of the four gametes descended from cell X will be haploid, and two will be diploid.

One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called?
A) deletion
B) transversion

C) inversion
D) translocation
E) duplication

A nonreciprocal crossover causes which of the following products?

A) deletion only

B) duplication only
C) nondisjunction

D) deletion and duplication
E) duplication and nondisjunction

Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual?
A) 47, +21
B) 47, XXY

C) 47, XXX
D) 47, XYY
E) 45, X

A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12. Although he is normal because his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm?
A) One-fourth will be normal, 1/4 will have the translocation, and 1/2 will have duplications and deletions.
B) All will carry the same translocation as the father.
C) None will carry the translocation because abnormal sperm will die.
D) His sperm will be sterile and the couple might consider adoption.
E) One-half will be normal and the rest will have the father's translocation.

Abnormal chromosomes are frequently found in malignant tumors. Errors such as translocations may place a gene in close proximity to different control regions. Which of the following might then occur to make the cancer worse?
A) an increase in nondisjunction

B) expression of inappropriate gene products C) a decrease in mitotic frequency
D) death of the cancer cells in the tumor
E) sensitivity of the immune system

An inversion in a human chromosome often results in no demonstrable phenotypic effect in the individual. What else may occur?
A) There may be deletions later in life.
B) Some abnormal gametes may be formed.

C) There is an increased frequency of mutation.

D) All inverted chromosomes are deleted.
E) The individual is more likely to get cancer.

What is the source of the extra chromosome 21 in an individual with Down syndrome?

A) nondisjunction in the mother only
B) nondisjunction in the father only
C) duplication of the chromosome

D) nondisjunction or translocation in either parent

E) It is impossible to detect with current technology.

Down syndrome has a frequency in the U.S. population of ~1/700 live births. In which of the following groups would you expect this frequency to be significantly higher?
A) people in Latin or South America
B) the Inuit and other peoples in very cold habitats

C) people living in equatorial areas of the world

D) very small population groups
E) No groups have such higher frequency.

A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition?
A) The woman inherited this tendency from her parents.
B) One member of the couple carried a translocation.

C) One member of the couple underwent nondisjunction in somatic cell production.

D) One member of the couple underwent nondisjunction in gamete production.
E) The mother had a chromosomal duplication.

What is a syndrome?
A) a characteristic facial appearance
B) a group of traits, all of which must be present if an aneuploidy is to be diagnosed
C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation
D) a characteristic trait usually given the discoverer's name
E) a characteristic that only appears in conjunction with one specific aneuploidy

Which of the following is known as a Philadelphia chromosome?
A) a human chromosome 22 that has had a specific translocation
B) a human chromosome 9 that is found only in one type of cancer
C) an animal chromosome found primarily in the mid-Atlantic area of the United States

D) an imprinted chromosome that always comes from the mother

E) a chromosome found not in the nucleus but in mitochondria

At what point in cell division is a chromosome lost so that, after fertilization with a normal gamete, the result is an embryo with 45, X?

I. an error in anaphase I
II. an error in anaphase II
III. an error of the first postfertilization mitosis IV.an error in pairing
A) I or II only
B) II or IV only
C) III or IV only
D) I, II, or III only
E) I, II, III, or IV

Which of the following is true of aneuploidies in general?
A) A monosomy is more frequent than a trisomy.
B) 45, X is the only known human live-born monosomy.
C) Some human aneuploidies have selective advantage in some environments.
D) Of all human aneuploidies, only Down syndrome is associated with mental retardation.

E) An aneuploidy resulting in the deletion of a chromosome segment is less serious than a duplication.

A woman is found to have 47 chromosomes, including three X chromosomes. Which of the following describes her expected phenotype?
A) masculine characteristics such as facial hair
B) enlarged genital structures

C) excessive emotional instability
D) healthy female of slightly above-average height

E) sterile female

If recombination frequency is equal to distance in map units, what is the approximate distance between genes A and B?
A) 1.5 map units
B) 3 map units

C) 6 map units
D) 15 map units
E) 30 map units

What is the greatest benefit of having used a testcross for this experiment?
A) The homozygous recessive parents are obvious to the naked eye.
B) The homozygous parents are the only ones whose crossovers make a difference. C) Progeny can be scored by their phenotypes alone.
D) All of the progeny will be heterozygous.
E) The homozygous recessive parents will be unable to cross over.

How many of their daughters might be expected to be color-blind dwarfs?

A) all
B) none
C) half

D) one out of four
E) three out of four

What proportion of their sons would be color-blind and of normal height?

A) none
B) half
C) one out of four

D) three out of four
E) all

They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes?
A) 0%
B) 25%

C) 50%
D) 75%
E) 100%

In his transformation experiments, what did Griffith observe?
A) Mutant mice were resistant to bacterial infections.
B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.
C) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic.
D) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.
E) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.

How do we describe transformation in bacteria?
A) the creation of a strand of DNA from an RNA molecule

B) the creation of a strand of RNA from a DNA molecule

C) the infection of cells by a phage DNA molecule
D) the type of semiconservative replication shown by DNA

E) assimilation of external DNA into a cell

After mixing a heat-killed, phosphorescent (light-emitting) strain of bacteria with a living, nonphosphorescent strain, you discover that some of the living cells are now phosphorescent. Which observation(s) would provide the best evidence that the ability to phosphoresce is a heritable trait?

A) DNA passed from the heat-killed strain to the living strain.
B) Protein passed from the heat-killed strain to the living strain.
C) The phosphorescence in the living strain is especially bright.
D) Descendants of the living cells are also phosphorescent.
E) Both DNA and protein passed from the heat-killed strain to the living strain.

In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts?
A) DNA contains sulfur, whereas protein does not.
B) DNA contains phosphorus, whereas protein does not.

C) DNA contains nitrogen, whereas protein does not.
D) DNA contains purines, whereas protein includes pyrimidines.
E) RNA includes ribose, whereas DNA includes deoxyribose sugars.

Which of the following investigators was (were) responsible for the following discovery?
In DNA from any species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine.
A) Frederick Griffith
B) Alfred Hershey and Martha Chase
C) Oswald Avery, Maclyn McCarty, and Colin MacLeod
D) Erwin Chargaff
E) Matthew Meselson and Franklin Stahl

Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine?

A) 8%
B) 16%

C) 31%
D) 42%
E) It cannot be determined from the information provided.

Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA?
A) the diameter of the helix
B) the rate of replication

C) the sequence of nucleotides
D) the bond angles of the subunits
E) the frequency of A vs. T nucleotides

It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following?

A) sequence of bases
B) phosphate-sugar backbones

C) complementary pairing of bases

D) side groups of nitrogenous bases

E) different five-carbon sugars

In an analysis of the nucleotide composition of DNA, which of the following will be found? A) A = C
B) A = G and C = T
C) A + C = G + T

D) G + C = T + A

What is meant by the description "antiparallel" regarding the strands that make up DNA?

A) The twisting nature of DNA creates nonparallel strands.
B) The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.

C) Base pairings create unequal spacing between the two DNA strands.

D) One strand is positively charged and the other is negatively charged.
E) One strand contains only purines and the other contains only pyrimidines.

Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons?
A) Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not.
B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many.

C) The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes.

D) Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not.

E) Prokaryotes have telomeres, and eukaryotes do not.

Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base?
A) One of the daughter cells, but not the other, would have radioactive DNA.
B) Neither of the two daughter cells would be radioactive.
C) All four bases of the DNA would be radioactive.
D) Radioactive thymine would pair with nonradioactive guanine.
E) DNA in both daughter cells would be radioactive.

An Okazaki fragment has which of the following arrangements?

A) primase, polymerase, ligase
B) 3' RNA nucleotides, DNA nucleotides 5'
C) 5' RNA nucleotides, DNA nucleotides 3'

D) DNA polymerase I, DNA polymerase III

E) 5' DNA to 3'

In E. coli, there is a mutation in a gene called dnaB that alters the helicase that normally acts at the origin. Which of the following would you expect as a result of this mutation?
A) No proofreading will occur.
B) No replication fork will be formed.

C) The DNA will supercoil.
D) Replication will occur via RNA polymerase alone.
E) Replication will require a DNA template from another source.

Which enzyme catalyzes the elongation of a DNA strand in the 5' → 3' direction?

A) primase
B) DNA ligase
C) DNA polymerase III

D) topoisomerase
E) helicase

At a specific area of a chromosome, the following sequence of nucleotides is present where the chain opens to form a replication fork:
3' C C T A G G C T G C A A T C C 5'
An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence?

A) 5' G C C T A G G 3'

B) 3' G C C T A G G 5'

C) 5' A C G T T A G G 3'

D) 5' A C G U U A G G 3'

E) 5' G C C U A G G 3'

Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA strands that are aligned in parallel arrays. How could these arise?
A) replication followed by mitosis
B) replication without separation

C) meiosis followed by mitosis
D) fertilization by multiple sperm
E) special association with histone proteins

To repair a thymine dimer by nucleotide excision repair, in which order do the necessary enzymes act?
A) exonuclease, DNA polymerase III, RNA primase
B) helicase, DNA polymerase I, DNA ligase

C) DNA ligase, nuclease, helicase
D) DNA polymerase I, DNA polymerase III, DNA ligase

E) endonuclease, DNA polymerase I, DNA ligase

What is the function of DNA polymerase III?
A) to unwind the DNA helix during replication
B) to seal together the broken ends of DNA strands
C) to add nucleotides to the 3' end of a growing DNA strand

D) to degrade damaged DNA molecules

E) to rejoin the two DNA strands (one new and one old) after replication

The difference between ATP and the nucleoside triphosphates used during DNA synthesis is that
A) the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose.
B) the nucleoside triphosphates have two phosphate groups; ATP has three phosphate groups.

C) ATP contains three high-energy bonds; the nucleoside triphosphates have two.

D) ATP is found only in human cells; the nucleoside triphosphates are found in all animal and plant cells.
E) triphosphate monomers are active in the nucleoside triphosphates, but not in ATP.

The leading and the lagging strands differ in that
A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.
B) the leading strand is synthesized by adding nucleotides to the 3' end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5' end.
C) the lagging strand is synthesized continuously, whereas the leading strand is synthesized in short fragments that are ultimately stitched together.
D) the leading strand is synthesized at twice the rate of the lagging strand.

A new DNA strand elongates only in the 5' to 3' direction because
A) DNA polymerase begins adding nucleotides at the 5' end of the template.
B) Okazaki fragments prevent elongation in the 3' to 5' direction.
C) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end.

D) replication must progress toward the replication fork.
E) DNA polymerase can only add nucleotides to the free 3' end.

What is the function of topoisomerase?
A) relieving strain in the DNA ahead of the replication fork
B) elongating new DNA at a replication fork by adding nucleotides to the existing chain

C) adding methyl groups to bases of DNA
D) unwinding of the double helix
E) stabilizing single-stranded DNA at the replication fork

What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
A) It synthesizes RNA nucleotides to make a primer.
B) It catalyzes the lengthening of telomeres.

C) It joins Okazaki fragments together.
D) It unwinds the parental double helix.
E) It stabilizes the unwound parental DNA.

Which of the following help(s) to hold the DNA strands apart while they are being replicated?
A) primase
B) ligase

C) DNA polymerase
D) single-strand binding proteins

E) exonuclease

Individuals with the disorder xeroderma pigmentosum are hypersensitive to sunlight. This occurs because their cells are impaired in what way?
A) They cannot replicate DNA.
B) They cannot undergo mitosis.

C) They cannot exchange DNA with other cells.
D) They cannot repair thymine dimers.
E) They do not recombine homologous chromosomes during meiosis.

Use the list of choices below for the following questions:

1. helicase
2. DNA polymerase III
3. ligase
4. DNA polymerase I
5. primase Which of the enzymes synthesizes short segments of RNA? A) I B) II C) III D) IV E) V

Use the list of choices below for the following questions:

1. helicase
2. DNA polymerase III
3. ligase
4. DNA polymerase I
5. primase Which of the enzymes removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3' end of Okazaki fragments? A) I B) II C) III D) IV E) V

Use the list of choices below for the following questions:

1. helicase
2. DNA polymerase III
3. ligase
4. DNA polymerase I
5. primase Which of the enzymes separates the DNA strands during replication? A) I B) II C) III D) IV E) V

Use the list of choices below for the following questions:

1. helicase
2. DNA polymerase III
3. ligase
4. DNA polymerase I
5. primase Which of the enzymes covalently connects segments of DNA? A) I B) II C) III D) IV E) V

Given the damage caused by UV radiation, the kind of gene affected in those with XP is one whose product is involved with
A) mending of double-strand breaks in the DNA backbone.
B) breakage of cross-strand covalent bonds.

C) the ability to excise single-strand damage and replace it.

D) the removal of double-strand damaged areas.
E) causing affected skin cells to undergo apoptosis.

Which of the following sets of materials is required by both eukaryotes and prokaryotes for replication?
A) double-stranded DNA, four kinds of dNTPs, primers, origins of replication
B) topoisomerases, telomerases, polymerases

C) G-C rich regions, polymerases, chromosome nicks

D) nucleosome loosening, four dNTPs, four rNTPs

E) ligase, primers, nucleases

Studies of nucleosomes have shown that histones (except H1) exist in each nucleosome as two kinds of tetramers: one of 2 H2A molecules and 2 H2B molecules, and the other as 2 H3 and 2 H4 molecules. Which of the following is supported by this data?
A) DNA can wind itself around either of the two kinds of tetramers.
B) The two types of tetramers associate to form an octamer.
C) DNA has to associate with individual histones before they form tetramers.
D) Only H2A can form associations with DNA molecules.
E) The structure of H3 and H4 molecules is not basic like that of the other histones.

In a linear eukaryotic chromatin sample, which of the following strands is looped into domains by scaffolding?
A) DNA without attached histones
B) DNA with H1 only

C) the 10-nm chromatin fiber

D) the 30-nm chromatin fiber

E) the metaphase chromosome

Which of the following statements describes the eukaryotic chromosome?
A) It is composed of DNA alone.
B) The nucleosome is its most basic functional subunit.
C) The number of genes on each chromosome is different in different cell types of an organism.

D) It consists of a single linear molecule of double-stranded DNA plus proteins.

E) Active transcription occurs on heterochromatin but not euchromatin.

If a cell were unable to produce histone proteins, which of the following would be a likely effect?
A) There would be an increase in the amount of "satellite" DNA produced during centrifugation.

B) The cell's DNA couldn't be packed into its nucleus.

C) Spindle fibers would not form during prophase.
D) Amplification of other genes would compensate for the lack of histones.
E) Pseudogenes would be transcribed to compensate for the decreased protein in the cell.

Which of the following statements is true of histones?
A) Each nucleosome consists of two molecules of histone H1.
B) Histone H1 is not present in the nucleosome bead; instead, it draws the nucleosomes together.

C) The carboxyl end of each histone extends outward from the nucleosome and is called a "histone tail."
D) Histones are found in mammals, but not in other animals or in plants or fungi.
E) The mass of histone in chromatin is approximately nine times the mass of DNA.

Why do histones bind tightly to DNA?
A) Histones are positively charged, and DNA is negatively charged.

B) Histones are negatively charged, and DNA is positively charged.

C) Both histones and DNA are strongly hydrophobic.
D) Histones are covalently linked to the DNA.
E) Histones are highly hydrophobic, and DNA is hydrophilic.

Which of the following represents the order of increasingly higher levels of organization of chromatin?
A) nucleosome, 30-nm chromatin fiber, looped domain
B) looped domain, 30-nm chromatin fiber, nucleosome

C) looped domain, nucleosome, 30-nm chromatin fiber

D) nucleosome, looped domain, 30-nm chromatin fiber

E) 30-nm chromatin fiber, nucleosome, looped domain

Which of the following statements describes chromatin?
A) Heterochromatin is composed of DNA, whereas euchromatin is made of DNA and RNA.

B) Both heterochromatin and euchromatin are found in the cytoplasm.
C) Heterochromatin is highly condensed, whereas euchromatin is less compact.
D) Euchromatin is not transcribed, whereas heterochromatin is transcribed.
E) Only euchromatin is visible under the light microscope.

Which of the following modifications is least likely to alter the rate at which a DNA fragment moves through a gel during electrophoresis?
A) altering the nucleotide sequence of the DNA fragment without adding or removing nucleotides

B) acetylating the cytosine bases within the DNA fragment
C) increasing the length of the DNA fragment
D) decreasing the length of the DNA fragment
E) neutralizing the negative charges within the DNA fragment

Assume that you are trying to insert a gene into a plasmid. Someone gives you a preparation of genomic DNA that has been cut with restriction enzyme X. The gene you wish to insert has sites on both ends for cutting by restriction enzyme Y. You have a plasmid with a single site for Y, but not for X. Your strategy should be to

A) insert the fragments cut with restriction enzyme X directly into the plasmid without cutting the plasmid.
B) cut the plasmid with restriction enzyme X and insert the fragments cut with restriction enzyme Y into the plasmid.

C) cut the DNA again with restriction enzyme Y and insert these fragments into the plasmid cut with the same enzyme.
D) cut the plasmid twice with restriction enzyme Y and ligate the two fragments onto the ends of the DNA fragments cut with restriction enzyme X.

E) cut the plasmid with restriction enzyme X and then insert the gene into the plasmid.

How does a bacterial cell protect its own DNA from restriction enzymes?

A) by adding methyl groups to adenines and cytosines

B) by using DNA ligase to seal the bacterial DNA into a closed circle
C) by adding histones to protect the double-stranded DNA

D) by forming "sticky ends" of bacterial DNA to prevent the enzyme from attaching

E) by reinforcing the bacterial DNA structure with covalent phosphodiester bonds

What is the most logical sequence of steps for splicing foreign DNA into a plasmid and inserting the plasmid into a bacterium?

1. Transform bacteria with a recombinant DNA molecule.
2. Cut the plasmid DNA using restriction enzymes.
3. Extract plasmid DNA from bacterial cells.
4. Hydrogen-bond the plasmid DNA to nonplasmid DNA fragments.
5. Use ligase to seal plasmid DNA to nonplasmid DNA.

A) I, II, IV, III, V
B) II, III, V, IV, I
C) III, II, IV, V, I
D) III, IV, V, I, II
E) IV, V, I, II, III

Why is it so important to be able to amplify DNA fragments when studying genes?

A) DNA fragments are too small to use individually.
B) A gene may represent only a millionth of the cell's DNA.
C) Restriction enzymes cut DNA into fragments that are too small.

D) A clone requires multiple copies of each gene per clone.
E) It is important to have multiple copies of DNA in the case of laboratory error.

The reason for using Taq polymerase for PCR is that
A) it is heat stable and can withstand the heating step of PCR.
B) only minute amounts are needed for each cycle of PCR.
C) it binds more readily than other polymerases to the primers.
D) it has regions that are complementary to the primers.
E) it is heat stable, and it binds more readily than other polymerases to the primers.

Once the pattern found after one round of replication was observed, Meselson and Stahl could be confident of which of the following conclusions?
A) Replication is semi-conservative.
B) Replication is not dispersive.

C) Replication is not semi-conservative.
D) Replication is not conservative.
E) Replication is neither dispersive nor conservative.

In an experiment, DNA is allowed to replicate in an environment with all necessary enzymes,

dATP, dCTP, dGTP, and radioactively labeled dTTP (3H thymidine). After several minutes, the DNA is switched to nonradioactive medium and is then viewed by electron microscopy and autoradiography. Figure 13.2 represents the results. It shows a replication bubble, and the dots represent radioactive material. Which of the following is the most likely interpretation of the results?

A) There are two replication forks going in opposite directions.
B) Thymidine is being added only where the DNA strands are farthest apart.

C) Thymidine is being added only at the very beginning of replication.
D) Replication proceeds in one direction only.

Which enzyme was used to produce the molecule in Figure 13.3?

A) ligase
B) transcriptase
C) a restriction enzyme

D) RNA polymerase
E) DNA polymerase

For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work?

A) There is no radioactive isotope of nitrogen.
B) Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.
C) Avery et al. have already concluded that this experiment showed inconclusive results.
D) Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive.
E) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

You briefly expose bacteria undergoing DNA replication to radioactively labeled nucleotides. When you centrifuge the DNA isolated from the bacteria, the DNA separates into two classes. One class of labeled DNA includes very large molecules (thousands or even millions of nucleotides long), and the other includes short stretches of DNA (several hundred to a few thousand nucleotides in length). These two classes of DNA probably represent

A) leading strands and Okazaki fragments.

B) lagging strands and Okazaki fragments.

C) Okazaki fragments and RNA primers.

D) leading strands and RNA primers.

E) RNA primers and mitochondrial DNA.

Bacteria that contain the plasmid, but not the eukaryotic gene, would grow
A) in the nutrient broth plus ampicillin, but not in the broth containing tetracycline.

B) only in the broth containing both antibiotics.
C) in the broth containing tetracycline, but not in the broth containing ampicillin.

D) in all four types of broth.
E) in the nutrient broth without antibiotics only.

Bacteria containing a plasmid into which the eukaryotic gene has integrated would grow A) in the nutrient broth only.
B) in the nutrient broth and the tetracycline broth only.
C) in the nutrient broth, the ampicillin broth, and the tetracycline broth.

D) in all four types of broth.
E) in the ampicillin broth and the nutrient broth.

Bacteria that do not take up any plasmids would grow on which media?

A) the nutrient broth only
B) the nutrient broth and the tetracycline broth
C) the nutrient broth and the ampicillin broth

D) the tetracycline broth and the ampicillin broth

E) all three broths

Why might they be conducting such an experiment?
A) to find the location of this gene in the human genome
B) to prepare to isolate the chromosome on which the gene of interest is found

C) to find which of the students has which alleles
D) to collect population data that can be used to assess natural selection
E) to collect population data that can be used to study genetic drift

Analysis of the data obtained shows that two students each have two fragments, two students each have three fragments, and two students each have one only. What does this demonstrate?

A) Each pair of students has a different gene for this function.
B) The two students who have two fragments have one restriction site in this region.

C) The two students who have two fragments have two restriction sites within this gene.

D) The students with three fragments are said to have "fragile sites."
E) Each of these students is heterozygous for this gene.

In his work with pneumonia-causing bacteria and mice, Griffith found that
A) the protein coat from pathogenic cells was able to transform nonpathogenic cells.
B) heat-killed pathogenic cells caused pneumonia.
C) some substance from pathogenic cells was transferred to nonpathogenic cells, making them pathogenic.
D) the polysaccharide coat of bacteria caused pneumonia.
E) bacteriophages injected DNA into bacteria

What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
A) The origins of replication occur only at the 5' end.
B) Helicases and single-strand binding proteins work at the 5' end.

C) DNA polymerase can join new nucleotides only to the 3' end of a growing strand.

D) DNA ligase works only in the 3' → 5' direction.
E) Polymerase can work on only one strand at a time.

In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules?
A) A = G
B) A + G = C + T

C) A + T = G + T
D) A = C
E) G = T

The elongation of the leading strand during DNA synthesis

A) progresses away from the replication fork.
B) occurs in the 3' → 5' direction.
C) produces Okazaki fragments.

D) depends on the action of DNA polymerase.

E) does not require a template strand.

In a nucleosome, the DNA is wrapped around A) polymerase molecules.
B) ribosomes.
C) histones.

D) a thymine dimer.
E) satellite DNA.

E. coli cells grown on 15N medium are transferred to 14N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?
A) one high-density and one low-density band

B) one intermediate-density band
C) one high-density and one intermediate-density band

D) one low-density and one intermediate-density band

E) one low-density band

A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture?

A) DNA polymerase

B) DNA ligase
C) nucleotides
D) Okazaki fragments

E) primase

The spontaneous loss of amino groups from adenine in DNA results in hypoxanthine, an uncommon base, opposite thymine. What combination of proteins could repair such damage?

A) nuclease, DNA polymerase, DNA ligase
B) topoisomerase, primase, DNA polymerase

C) topoisomerase, helicase, single-strand binding protein

D) DNA ligase, replication fork proteins, adenylyl cyclase

E) nuclease, topoisomerase, primase