Genetics Test 3

-ribonucleoprotein

A molecule that possesses both a protein component and an RNA component is referred to as a ________. -holoenzyme -tetramer -dimer -polymerase -ribonucleoprotein

Both deletions and duplications may be due to unequal crossing over

Which of these may be due to unequal crossing over? -Deletion -Duplication -Insertion -Nonreciprocal translocation -Both deletions and duplications may be due to unequal crossing over

the pseudoautosomal regions

Which region(s) of the X chromosome recombine during meiosis? -the heterochromain -the pseudoautosomal regions -the centromere -SRY -the male-specific region of the Y

Individuals with a Y chromosome are male.

Which statement best describes human sex determination? -Females are heterogametic. -Individuals with a Y chromosome are male. -Individuals with two X chromosomes are female. -Individuals with at least twice as many X chromosomes as Y chromosomes are female. -Individuals with one X chromosome are male.

FISH

Which technique can be used to identify the location of genes on a chromosome? X-ray diffraction analysis electrophoresis reassociation kinetics FISH

gene and mRNA; a ribosome and a tRNA

A particular ________ carry the information for making a particular protein, but ________ can be used to make any protein. -gene and tRNA; a ribosome and an mRNA -ribosome and mRNA; a gene and a tRNA -gene and mRNA; a ribosome and a tRNA -gene and ribosome; a tRNA and an mRNA -tRNA and ribosome; a gene and an mRNA

SRY

A small part of the human Y chromosome contains the gene that is responsible for determining maleness. What is the name of this gene? medulla Dax NRY SRY Xist

paracentric

A ________ inversion is one whose breakpoints do not flank the centromere. dicentric acentric paracentric segmental pericentric

nondisjunction in either meiosis I or meiosis II of the mother

A child is born with Turner's syndrome and she is red green color blind (recessive X-linked). Her father is red green color blind as well and her mother is homozygous dominant for color sight. What had to happen in order for this child to be born with this chromosomal complement? -nondisjunction in meiosis II of the mother -nondisjunction in either meiosis I or meiosis II of the father -nondisjunction in meiosis I of the father -nondisjunction in either meiosis I or meiosis II of the mother -nondisjunction in meiosis II of the father

acentric

A chromosome without a centromere is ________. paracentric dicentric segmental acentric pericentric

nondisjunction in the other in meiosis I or II

A color-blind woman with Turner syndrome (XO) has a father who is color blind. Given that the gene for the color-blind condition is recessive and X-linked, provide a likely explanation for the origin of the color-blind and cytogenetic conditions in the woman. -nondisjunction in the other in meiosis I or II -nondisjunction in the father at meiosis I -nondisjunction in the father in meiosis I or II -nondisjunction in the mother in meiosis I and in the father in meiosis II -no abnormalities had to occur to produce this daughter

various lengths of trinucleotide repeats

A genomic condition that may be responsible for some forms of fragile X syndrome, as well as Huntington disease, involves ________. -plasmids inserted into the FMR-1 gene -multiple inversions in the X chromosome -various lengths of trinucleotide repeats -single translocations in the X chromosome -multiple breakpoints fairly evenly dispersed along the X chromosome

trisomy

A human individual with 47 chromosomes has a _____. monosomy trisomy monoploidy diploidy triploidy

2

A human with the karyotype 49, XXXYY forms _______ Barr bodies. View Available Hint(s) 2 3 0 1

SRY

A karyotype is done on a phenotypically female patient, and it is discovered that she has a Y chromosome. However, the Y chromosome appears to have a deletion. What region is most likely deleted? PAR SRY centromere MSY heterochromatin of M S Y

c.47 XXY

A man observed his stained epithelial cells and saw a Barr body. If a karyotype was performed on his cells, what would you expect to see? a.46, XY b.46 XX c.47 XXY d.47 XYY e.Either c or d

Father Klinefelter's men are X X Y. The son must have received an X from his mother with the color blind allele on it. However, since he has color vision, his other X must contain the allele for color vision, which could only come from his father. So, there must have been nondisjunction in the father.

A man with Klinefelter's syndrome has normal color vision. His mother was colorblind, but his father had normal color vision. From this information, can you determine which parent underwent nondisjunction? -Mother -Father -Either could have undergone nondisjunction to end up with this result -Neither, as Klinefelter's syndrome is not due to nondisjunction -Not enough information to make a determination

nondisjunction in meiosis II of the mother

A son is born with Kleinfelter's syndrome and hemophilia. His father was normal and his mother was a carrier for the recessive X-linked blood clotting disorder. What occurred in meiosis to produce this genetic outcome? -nondisjunction in meiosis II of the mother -nondisjunction in meiosis I of the mother -nondisjunction in meiosis I of the father -nondisjunction in either meiosis I or meiosis II of the mother -nondisjunction in meiosis II of the father

XAXa, XAY

A young boy is diagnosed with hemophilia A, a recessive, X-linked condition. If neither parent has hemophilia A, what are their genotypes? a.XAXa, XAY b.XAXA, XaY c.XaXa, XAYA d.Mother could be either XAXA orXAXa, but father is XaY e.cannot predict from the information given

the conservative model was conclusively proven to be an inaccurate model

After ONE round of DNA replication in Meselson and Stahl's classic experiment, ________. -both the conservative and the semiconservative models were still possibilities -all three models of replication were still possibilities -the conservative model was conclusively proven to be an inaccurate model -the dispersive model was conclusively proven to be an inaccurate model -the semiconservative model was conclusively proven to be the accurate model

the ability to form a replication fork

All EXCEPT which of the following are differences between eukaryotic and prokaryotic DNA replication? -the rate of DNA synthesis -the number of replication origins -the ability to form a replication fork -the type and number of polymerases involved in DNA synthesis

The number of Barr bodies present. Barr bodies are unrelated to sex determination. They are inactivated X chromosomes.

All EXCEPT which of the following are mechanisms by which sex can be determined? -The presence of two X chromosomes or an X and a Y chromosome. -The ratio of X chromosomes to the number of sets of autosomes. -The number of Barr bodies present. -The temperature at which the eggs are incubated.

found in eukaryotes and prokaryotes

All EXCEPT which of the following are related to telomeres? -telomerase enzyme -found in eukaryotes and prokaryotes -links to the aging process -short tandem repeats located at the ends of telomeres

DNA has four nucleotides.

All EXCEPT which of the following statements are evidence that DNA, and not protein, is the genetic material in eukaryotes as well as bacteria? -UV light is most mutagenic at a wavelength at which DNA and RNA strongly absorb. -Introduction of a cloned DNA into another organism results in the production of the corresponding protein product. -DNA is located only where the primary genetic function is known to occur. -DNA has four nucleotides.

have 5′ to 3′ polymerization activity All known bacterial D N A polymerases extend D N A chains in the 5′ to 3′ direction. Some have 5′ to 3′ exonuclease activity, but none have 3′ to 5′ polymerization activity or the ability to initiate chain synthesis.

All known bacterial D N A polymerases _____. -can initiate D N A chain synthesis -have 5′ to 3′ polymerization activity -have 5′ to 3′ exonuclease activity -have 3′ to 5′ polymerization activity -all of the above

two haploid gametes fuse to form a diploid cell

All of the following events occur during normal meiosis except _______. -sister chromatids separate from one another during meiosis II -two haploid gametes fuse to form a diploid cell -homologous chromosomes separate from one another during meiosis I -one diploid cell produces four haploid cells

The figure shows three chromosomes.

Although triple-X human females typically have normal offspring, what kinds of gametes, with respect to the X chromosomes, would you expect from such XXX females? Which of the following is an example of metaphase of meiosis I? -The figure shows two chromatids. -The figure shows one chromosome. -The figure shows one chromatid. -The figure shows three chromosomes. -The figure shows two chromosomes.

-autotetraploid

An ________ may arise when three sperm cells are involved in fertilization of a single egg. -allotriploid -autotetraploid -autotriploid -aneuploidy -allotetraploid

allotetraploid, fertile An amphidiploid is a type of allotetraploid. Amphidiploids and allotetraploids both have four sets of haploid genomes from two different species. However, if we know the parental species origins of allotetraploids, we call them amphidiploids. Because they are 4n (2 × 2n), homologs can pair at meiosis and produce functional gametes.

An amphidiploid is a type of ___________ and is usually ___________. -autopolyploid, sterile -autopolyploid, fertile -allodiploid, sterile -allotetraploid, fertile -chromosomal abnormality, sterile

triploid

An expected meiotic pairing configuration in a ________ would be a trivalent. monosomic individual diploid trisomic individual triploid monoploid

said to have a trisomy

An individual with Patau syndrome would be ________. said to have a trisomy termed a triploid termed a haploid termed a euploid said to have a monosomy

the ability to directly influence the development of traits

Any molecule that serves as the genetic material must have the following characteristics except _____. -the ability to be replicated -the ability to store information -the ability to directly influence the development of traits -the ability to express information -the potential to be changed via mutation

complementary; antiparallel

Any two nucleic acid fragments will spontaneously hybridize if they encounter each other as long as they are ________ and ________. -identical; compatible -complementary; antiparallel -primed; replicated -transcribed; translated -DNA; RNA

mRNA, tRNA, rRNA

Arrange the principal forms of RNA in order of abundance from LEAST abundant to MOST abundant. rRNA, tRNA, mRNA tRNA, mRNA, rRNA mRNA, tRNA, rRNA mRNA, rRNA, tRNA tRNA, rRNA, mRNA

supercoiling; DNA gyrase

As unwinding of the helix occurs during DNA replication, tension referred to as ________ is created ahead of the replication fork. This tension is relieved by the action of ________. -supercoiling; DNA helicase -processivity; primase -processivity; DNA gyrase -supercoiling; single-stranded binding proteins -supercoiling; DNA gyrase

aneuploidy and trisomy

Assume that a species has a diploid chromosome number of 24. The term applied to an individual with 25 chromosomes would be ________. aneuploidy euploid triploid trisomy aneuploidy and trisomy

triploid

Assume that a species has a diploid chromosome number of 24. The term applied to an individual with 36 chromosomes would be ________. aneuploidy triploid tetraploid monoploid allopolyploid

28

Assume that an organism has a diploid chromosome number of 14. There would be ________ chromosomes in a tetraploid. 14 104 28 42 56

7

Assume that an organism has a haploid chromosome number of 7. There would be ________ chromosomes in a monoploid individual of that species. 28 7 42 14 3.5

No, the chromosomal basis for sex determination in Drosophila based on the balance between the number of X chromosomes and haploid sets of autosomes.

Assume that you are told that a particular organism, Drosophila, has the XO chromosome complement. You are also told that the autosomal complement is a normal 2n. You know that in humans the XO complement is female determining. Would you be correct in assuming that the Drosophila sex for XO is also female? Choose the answer that includes the correct explanation. -No, sex determination in Drosophila is dependent on the presence or absence of the Y chromosome. -Yes, because sex determination in humans and insects is essentially the same. -No, the chromosomal basis for sex determination in Drosophila based on the balance between the number of X chromosomes and haploid sets of autosomes. -No, it takes two X chromosomes to produce a female in humans and a Y chromosome to produce a male in Drosophila. -Yes, the presence of an X chromosome determines a female in both organisms.

all of the above

Autotriploids _____. -can result from the fertilization of a haploid gamete with an abnormal diploid gamete -can result from the fertilization of an egg by two haploid sperm -can result from the crossing of a diploid with a tetraploid -have three chromosome sets that are identical in terms of number and content -all of the above

Transformation is the process whereby one organism is genetically altered by exposure to DNA from another organism. Since DNA can carry heritable "traits" from one organism to another, it must be the genetic material.

Briefly define transformation and describe the relationship between the phenomenon of transformation and the discovery that DNA is the genetic material in bacteria. -Transformation is the process whereby RNA serves as a template for the synthesis of the complementary DNA molecule. Since DNA can be synthesized on the RNA molecule as a template, DNA must be the genetic material. -Transformation is the process whereby one organism is genetically altered by exposure to DNA from another organism. Since DNA can carry heritable "traits" from one organism to another, it must be the genetic material. -Transformation is the process whereby one organism transfers DNA to another organism through a direct contact. Since DNA can be transferred in such a way, it must be the genetic material. -Transformation is the process of infection of an organism by a viral nucleic acid. Since DNA is a nucleic acid, it must be the genetic material.

Yes, if the person is a translocation carrier. This rare event can occur if there is a translocation of most of an entire chromosome (shown here as a 14/21 translocation), then there is no loss of information, even if there are only 45 physically separate chromosomes.

Can a person have 45 chromosomes and be phenotypically normal? -Yes, this happens often. -Yes, if the person is a translocation carrier. -Yes, if it is the X chromosome that is monosomic. -No, not having 46 chromosomes will always result in some phenotype. -No, having only 45 chromosomes is lethal in humans.

Since females normally have two X chromosomes, random X inactivation during development of a heterozygous female will create a mottled pattern. Normal males are not mottled, because they only have one X chromosome and therefore can only express one of the two potential X-linked alleles.

Cat breeders are aware that kittens expressing the X-linked calico coat pattern and tortoiseshell pattern are almost invariably females. Which one of the following statements supports this conclusion? -Since females normally have two X chromosomes, expression of both potential X-linked alleles at the same time in all cells will create a mottled pattern. Normal males are not mottled, because they only have one X chromosome and therefore can only express one of the two potential X-linked alleles. -Since females normally have two X chromosomes, random X inactivation during development of a heterozygous female will create a mottled pattern. Normal males are not mottled, because they only have one X chromosome and therefore can only express one of the two potential X-linked alleles. -Since males normally have one X and one Y chromosome, random X inactivation during development of a male will create a mottled pattern. Normal females are mottled, because they have two X chromosomes. -Since females normally have two X chromosomes, random X inactivation during development of a homozygous female will create a mottled pattern. Normal males are not mottled, because they are heterozygous and therefore express each one of the two potential X-linked alleles.

autopolyploids

Certain varieties of chrysanthemums contain 18, 36, 54, 72, and 90 chromosomes; all are multiples of a basic set of nine chromosomes. How would you describe these varieties genetically? aneuploids autopolyploids allopolyploids amphipolyploids

more; fewer; right-handed

Compared with B-DNA, A-DNA is ________ compact, has ________ base pairs per turn, and is a ________ helix. more; fewer; right-handed less; more; right-handed less; fewer; left-handed more; more; right-handed more; fewer; left-handed

initiates at multiple origins

DNA replication in eukaryotes ________. -initiates at multiple origins -synthesizes DNA approximately 25 times faster than in prokaryotes -occurs without the need of a primer -utilizes a single type of DNA polymerase -takes place multiple times per cell cycle

semiconservatively

DNA replication proceeds ________. dispersively semiconservatively discontinuously unidirectionally progressively

in the place of the primer RNA after it is removed

DNA polymerase I is thought to add nucleotides ________. -to the 5' end of the primer -on single-stranded templates without need for an RNA primer -to the 3' end of the primer -in a 5' to 5' direction -in the place of the primer RNA after it is removed

to the 3' end of the RNA primer

DNA polymerase III adds nucleotides ________. -in the place of the primer RNA after it is removed -to the 3' end of the RNA primer -to both ends of the RNA primer -to the 5' end of the RNA primer -to internal sites in the DNA template

Inactivation of an X chromosome occurs at random early in embryonic development. Inactivated X chromosomes are in some way "marked" such that all clonally related cells have the same X chromosome inactivated.

Define the Lyon hypothesis. -Inactivation of a particular X chromosome depends on its genetic content and occurs after differentiation of cell type. Different cell types may inactivate different X chromosomes. -Inactivation of a particular X chromosome depends on its genetic content. Inactivated X chromosomes are in some way "marked" such that all clonally related cells have the same X chromosome inactivated. -Inactivation of an X chromosome occurs at random just after fertilization. Inactivated X chromosomes are in some way "marked" such that all clonally related cells have the same X chromosome inactivated. -Inactivation of an X chromosome occurs at random early in embryonic development. Inactivated X chromosomes are in some way "marked" such that all clonally related cells have the same X chromosome inactivated.

an X chromosome with two duplications of the 16A region

Describe double Bar mutations in Drosophila melanogaster. -a female fly with only one X chromosome -an inversion heterozygote of the X chromosome -large deletion in the X chromosome -duplications in a portion of the X chromosome -an X chromosome with two duplications of the 16A region

3. Growing the cells for many generations in medium containing 15N and transferring to 14N medium allows for the detection of new DNA. 5. Labeling the pool of nitrogenous bases of the DNA of E. coli with the heavy isotope 15N labels allows old DNA to be followed. 6. A comparison of the density of DNA samples at various times in the experiment allows an estimation of amounts of old and new DNA .

Describe the role of 15N in the Meselson-Stahl experiment. Select the three correct statements. 1. Medium containing 15N is more nutritious, so growing the cells in such medium allows results to be obtained in a shorter time. 2. Labeling the pool of nitrogenous bases of the RNA with 15N allows for the detection of newly synthesized DNA. 3. Growing the cells for many generations in medium containing 15N and transferring to 14N medium allows for the detection of new DNA. 4. Electrophoresis enables the separation of labeled DNA and estimates the amounts of DNA at various times during the experiment. 5. Labeling the pool of nitrogenous bases of the DNA of E. coli with the heavy isotope 15N labels allows old DNA to be followed. 6. A comparison of the density of DNA samples at various times in the experiment allows an estimation of amounts of old and new DNA .

amphidiploid

Doubling the chromosomes of a sterile species hybrid with colchicine or cold shock is a method used to produce a fertile species hybrid called a ________. autoallopolyploid amphidiploid triploid allopolyploid autoploid

male

Drosophila, like humans, have X and Y chromosomes. However, unlike humans, it is the ratio of the number of X chromosomes to sets of autosomes that determines the sex of the fly. A fly has one X chromosome, two sets of autosomes (a diploid set), and no Y. What is its sex? a.male b.female c.intersex d.metamale e.metafemale

Okazaki fragments; lagging; bidirectionality

During DNA replication, ________ are 1000-2000 nucleotide long strands synthesized on the ________ strand to maintain the ________ of replication. -primers; leading; discontinuity -primers; lagging; continuity -Okazaki fragments; leading; semidiscontinuity -Okazaki fragments; lagging; bidirectionality -Okazaki fragments; leading; accuracy

A mixture of different sized DNA molecules are subjected to an electric field, and migrate through a semisolid matrix such that smaller fragments travel further in the matrix.

Explain the technique of electrophoretic separation of DNA fragments. -A mixture of different sized DNA molecules are subjected to an electric field, and then to a magnetic field. The fragments migrate through a liquid matrix such that smaller fragments travel further in the matrix. -A mixture of different sized DNA molecules are subjected to an electric field, and migrate through a solid matrix such that larger fragments travel further in the matrix. -A mixture of different sized DNA molecules are subjected to an electric field, and migrate through a semisolid matrix such that smaller fragments travel further in the matrix. -A mixture of different sized DNA molecules are subjected to an electric field, and then to a magnetic field. The fragments migrate through a semisolid matrix such that larger fragments travel further in the matrix.

DNA is found where primary genetic functions occur; protein is found everywhere.

Explain why the distribution of protein and nucleic acid in a eukaryotic cell favors nucleic acid as the genetic material. -DNA is found in the nucleus; protein is found in the cytoplasm. -DNA is found everywhere; protein is found only near the localisation of ribosomes. -DNA is found everywhere; protein is found where primary genetic functions occur. -DNA is found where primary genetic functions occur; protein is found everywhere.

chromosomal location of specific genetic information

FISH can be used to determine the ________. -difference in lengths between two fragments of DNA -chromosomal location of specific genetic information -locations of major and minor grooves in a double helix -difference between A-form and B-form DNAs -melting temperature of a fragment of DNA

homogametic

In Drosophila, the female is the ________ sex. isogametic heterogametic mosaic homogametic genic balance

one

For an individual with the XXY chromosomal composition, the expected number of Barr bodies in interphase cells is ________. three two one variable zero

are often associated with regions of short tandem D N A sequence repeats The expansion of the number of trinucleotide tandem repeats at a site on the X chromosome weakens that site of the chromosome, probably because the chromatin does not tightly coil, leading to fragile-X syndrome.

Fragile sites _____. -are always associated with a phenotypic disorder -are limited to the X chromosome -have been implicated in most forms of cancer -are often associated with regions of short tandem DNA sequence repeats -all of the above

a chemical component of cells could introduce a new, heritable trait to a cell

Frederick Griffith's transformation experiment showed that _____. -a chemical component of cells could introduce a new, heritable trait to a cell -D N A is the genetic material of viruses -D N A is the genetic material of bacteria -D N A is the genetic material of mice -all of the above

trinucleotide repeats increase in future generations

Genetic anticipation is best described as ________. -trinucleotide repeats form fragile sites -trinucleotide repeats increase in future generations -the phenomenon by which synapsis is delayed due to translocation cross formation -the G2 gap before meiosis -cells stop during the cell cycle to check for DNA damage or acentric fragments

reptiles

Genotypic sex determination does not in ________. mammals round worms birds reptiles flies

male; male; female

Give the sex of the following Drosophila assuming that the autosomes are present in the normal number. XO; XY; XXY. Answers are in order (respectively). female; male; male male; male; female male; male; male female; male; female male; female; female

69, 47

Given that a human normally contains 46 chromosomes, give the chromosome number for each of the following conditions: Triploid Trisomy 13 69, 47 69, 45 138, 47 96, 47 138, 45

45, 47

Given that a human normally contains 46 chromosomes, give the chromosome number for each of the following conditions: Turner syndrome (female, no Barr bodies) Klinefelter syndrome (male, one Barr body) 47, 45 45, 47 47, 47 46, 47 45, 46

the condition whereby a single chromosome is insufficient to sustain life

Haploinsufficiency refers to ________. -the condition whereby a single cell is insufficient to divide to cause cancer -a state of being whereby a single gene is sufficient to cause several phenotypes -the condition whereby a single chromosome is insufficient to sustain life -the genetic predisposition for some genes to come in only one copy in the genome -the process by which a single gene will cause a cascading effect on a genome's phenotype

euploid

Having a complete set or sets of chromosomes is called ________. aneuploidy ploidy diploid euploid monoploid

Uneven crossing over during meiotic prophase

How can duplications arise? -A chromosome break and exchange -Spindle failure during chromosome separation -Loss of a telomere -Uneven crossing over during meiotic prophase

as an amphidiploid between an Old World and a New World species of cotton plant

How did cultivated American cotton originate? -as an allotetraploid of two species of New World cotton plants -as an amphidiploid between an Old World and a New World species of cotton plant -from a Robertsonian translocation of a New World cotton plant -as an autotetraploid of Old World cotton plants -by gene duplication in Old World cotton plants

chromosomes will break and the sticky ends will rejoin

How do deletions and translocations in chromosomes often occur? -chromosomes will fuse telomere to telomere -when cells undergo meiosis II, chromosomes naturally break when sister chromatids are being pulled apart -chromosomes will break and the sticky ends will rejoin -during meiosis I, sister chromatid exchange leads to abnormalities -colchicine treatment causes chromosomal rearrangements

In females, one of the X chromosomes is condensed and largely inactive so that each sex has one active X chromosome.

How do mammals, including humans, solve the "dosage problem" caused by the presence of an X and Y chromosome in one sex and two X chromosomes in the other sex? -In females, one of the X chromosomes is condensed and largely inactive so that each sex has one active X chromosome. -In males, the X chromosome is up-regulated so that its genes are transcribed and translated twice as much as the X chromosomes in females. -Females produce twice the RNA transcripts as males do, but in females half of them are destroyed before being translated. -They don't, but the X chromosome contains genes whose expression is not quantitatively sensitive, so the differences are not important. -Male RNA transcripts from the X chromosome have leaders that cause them to be translated at twice the rate as those in females. -They don't; the fact that females have twice as much X chromosome protein products as males is critical for sex determination in mammals.

Diploid number of 5

How many chromosomes would be found in an allopolyploid plant if its parents had diploid numbers of 4 and 6 respectively? Haploid number of 5 Diploid number of 10 Diploid number of 5 Haploid number of 12

The clusters arose from the same ancestral gene.

Human adult hemoglobin is a tetramer containing two alpha (α) and two beta (β) polypeptide chains. The α gene cluster on chromosome 16 and the β gene cluster on chromosome 11 share amino acid similarities such that 61 of the amino acids of the α-globin polypeptide (141 amino acids long) are shared in identical sequence with the β-globin polypeptide (146 amino acids long). How might one explain the existence of two polypeptides with partially shared function and structure on two different chromosomes? -Similar genes arose by random chance. -The clusters arose from the same ancestral gene. -There is no explanation for the phenomenon yet.

The Taylor, Woods, and Hughes (1957) experiment used 3H-thymidine.

Identify a classic experiment that used autoradiography to determine the replicative nature of DNA in eukaryotes. -The Meselson-Stahl (1958) experiment used 15N-containing molecules. -The Kornberg (1957) experiment used four radiolabeled deoxyribonucleoside triphosphates (dNTPs). -The Okazaki (1968) experiment used small fragments containing RNA primers. -The Taylor, Woods, and Hughes (1957) experiment used 3H-thymidine.

deoxyribose, phosphate group, thymine

Identify three possible components of a DNA nucleotide. -deoxyribose, phosphate group, uracil -adenine, phosphate group, ribose -cytosine, phosphate group, ribose -guanine, phosphate group, ribose -cytidine, phosphate group, ribose -deoxyribose, phosphate group, thymine

Autoradiography is a technique that allows an isotope to be detected within a cell.

Identify what is meant by the term autoradiography. -Autoradiography is a technique that allows an isotope to be detected within a cell. -Autoradiography is a technique that use antibodies to detect structures inside a cell. -Autoradiography is a technique that allows to detect structures inside a cell with the use a fluorescent label. -Autoradiography is a technique that allows molecules of nucleic acids to be detected within a cell.

35%

If 15% of the nitrogenous bases in a sample of DNA from a particular organism are thymine, what percentage should be cytosine? 15% 35% 40% 30% 70%

1

If DNA replication were fully conservative, how many intact parental double helices would have been detected in Meselson and Stahl's experiment after three rounds of replication? 0 7 3 1 8

280 nm

If protein were the genetic principle and not nucleic acid, significant mutagenic effects would be detected at ________. 320 nm 280 nm 300 nm 260 nm 240 nm

As D N A is heated, the strands separate ("melt" or denature). An increase in single strands causes an increase in the A260. Once all the strands have separated, the A260 should plateau.

If you heated a sample of D N A from 30°Celsius to 100°Celsius while you measured the A260, which of the following graphs would you expect to see?

X chromosomes

In Drosophila, sex is determined by a balance between the number of haploid sets of autosomes and the number of ________. Y chromosomes X chromosomes centromeres nucleolar organizers telomeres

XO:2A

In Drosophila, the sex of a fly with the karyotype ________ is male. XX:2A XXY:2A XO:2A XYY:4A XXX:2A

Both b and d are correct Living smooth cells cause disease. However, mixing dead smooth cells with living rough ones transformed some rough cells into smooth which could cause disease.

In Griffin's transformation experiment, which strains of Streptococcus (called Diplococcus in his day) caused disease in mice? -Live I I R -Live I I I S -Dead I I I S -Live I I R mixed with dead I I I S -Both b and d are correct

sugar and phosphate

In a polynucleotide, the individual nucleotides are linked by bonds between _____. -nitrogenous bases -sugar and phosphate -sugars -phosphates -nitrogenous base and sugar

A + C = G + T

In an analysis of the nucleotide composition of double-stranded DNA to see which bases are equivalent in concentration, which of the following would be true? A + C = G + T A = C A = G and C = T A + T = G + C G = T

a portion of the Y chromosome

In humans, the genetic basis for determining the sex "male" is accomplished by the presence of ________. -high levels of estrogen -a portion of the Y chromosome -a balance between the number of X chromosomes and the number of haploid -one X chromosome -sets of autosomes

a. one old, heavy band and one new, light band After one round of replication there would be one set of original molecules in the "old" band built with the heavy isotope of 15N and a set of newly synthesized molecules in the "new" band, built with the light isotope 14N. The "old" band would remain intact after the second round of replication, and there would be more molecules in the "new" band.

In the Meselson-Stahl experiment, if D N A had been replicated conservatively, after two rounds of replication, there would be _____. -one old, heavy band and one new, light band -one "hybrid" band -one "hybrid" band and one new, light band -one "hybrid" band and one old, heavy band -one old, heavy band, one new, light band, and one "hybrid" band

conservative

In the Meselson-Stahl experiment, which mode of replication was eliminated based on data derived after one generation of replication? semiconservative conservative dispersive none of the modes

The bacteria were in the pellet, and many contained the radioactive viral DNA.

In the classic experiment conducted by Hershey and Chase, why was the pellet radioactive in the centrifuge tube that contained bacteria with viruses? -The bacteria were in the pellet, and they had incorporated radioactive proteins into their cell membranes. -The radioactive viruses were in the pellet, and the bacteria were in the supernatant. -The bacteria were in the pellet, and many contained the radioactive viral DNA. -The radioactive viruses (coats plus DNA) were in the pellet. -The radioactive protein coats of the viruses were in the pellet.

the synthesis of R N A from a D N A template

In the flow of information, transcription is _____. -the synthesis of R N A from a D N A template -the synthesis of protein from D N A directions -the synthesis of protein from R N A directions -the replication of D N A from a D N A template -the synthesis of D N A from an R N A template

two acentric fragments and two dicentric chromosomes

Inversion heterokaryotypes are often characterized as having reduced crossing over and reduced fertility. Assume that you were examining a strain of organisms you knew to be inversion heterokaryotypes and saw a relatively high number of double chromatid bridges extending between anaphase I nuclei, as indicated in the following drawing. What is the product of this type of inversion loop? -two chromosomes with deletions and two acentric fragments -four wild-type chromosomes -two acentric fragments and two dicentric chromosomes -two chromosomes with deletions and two with duplications -one acentric fragment, one dicentric chromosome, and two wild-type chromosomes

Yes, to generate primers. Primase can be thought of as a specialized R N A polymerase. It synthesizes short R N A primers from a D N A template that allow the formation of small, double-stranded D N A-R N A hybrid regions. These regions permit the initiation of D N A replication by D N A polymerase.

Is there any role for an R N A polymerase in D N A replication? -No, only D N A polymerases function in D N A replication. -Yes, in order to synthesize mRNA. -Yes, to generate primers. -Yes, to synthesize R N A that will bind to and keep the two D N A strands unwound. -Yes, in eukaryotes an R N A polymerase functions in telomere maintenance.

47, XYY

Jacobs syndrome in humans, which is manifested by a higher than average stature and potential behavioral problems, is caused by which chromosomal condition? 47, 21+ 47, XYY triploidy 47, XXY 45, X

autotetraploidy, assuming the normal chromosome complement is diploid

Name the polyploid condition that is formed from the addition of an extra set of chromosomes identical to the normal diploid complement of the same species. -allotetraploidy, assuming the normal chromosome complement is diploid -autooctoploidyploidy, assuming the normal chromosome complement is haploid -autotetraploidy, assuming the normal chromosome complement is haploid -autotetraploidy, assuming the normal chromosome complement is diploid -alloeuploidy, assuming the normal chromosome complement is diploid

amniocentesis and NIPGD

Name two methods used in genetic prenatal diagnostic testing in humans. -amniocentesis and fetal uterine physiology compatibility test -amniocentesis and DSCR -chorionic villus sampling and cellular mitotic analysis -amniocentesis and NIPGD -whole genome sequencing and PCR

either B or C

Nondisjunction in the first meiotic division in a male human could result at fertilization in _____. a.XYY condition b.Turner syndrome c.Klinefelter syndrome d.XXX syndrome e.either B or C

calico females, black females, yellow makes, and black males

Part complete Dosage compensation in mammals typically involves the random inactivation of one of the two X chromosomes relatively early in development. Such X chromosome inactivation often leads to phenotypic mosaicism. Assume that black fur in cats is due to the X-linked recessive gene b, whereas its dominant allele B produces yellow fur. A Bb heterozygote is a mosaic called "tortoise shell" or "calico." A mating between a black male and a calico female occurred. Give the phenotypes of the offspring. -all calico regardless of sex -calico females, black females, yellow makes, and black males -all black regardless of sex -calico females and yellow males -calico females, yellow females, black males, and yellow males

heterogametic

Part complete In humans, the male is the ________ sex. heteromorphic homogametic monogametic heterogametic homomorphic

it requires only a blood draw from the mother

Part complete Noninvasive prenatal genetic diagnosis is sometimes preferred to amniocentesis because ________. -the mother does not have to be present -it requires less technical time of a research lab -it gives more precise results -it gives quicker results -it requires only a blood draw from the mother

XX/XO scheme

Part complete The Protenor mode of sex determination is the ________. -scheme based on single translocations in the X chromosome -hermaphroditic scheme -XX/XO scheme -scheme based on F plasmids inserted into the FMR-1 gene -XO/YY scheme

a compensation loop

Part complete What is formed when meiosis occurs in an individual who is heterozygous for an intercalary deletion? -a compensation loop -a translocation cross -a terminal loop -a translocation -a trifecta chromosomal cross

Females will display mosaic retinas with patches of defective color perception and surrounding areas with normal color perception.

Predict the potential effect of the Lyon hypothesis on the retina of a human female heterozygous for the X-linked red-green color blindness trait. -X chromosome inactivation is influenced by methylation patterns inherited from the parents. When the chromosome with the X-linked red-green color blindness trait is obtained from the mother, then female offspring will not be affected. If it_s obtained from the father, then they will be affected. -Females will display mosaic retinas with patches of defective color perception and surrounding areas with normal color perception. -If descendants of the cells with the X-linked red-green color blindness trait participate in the further development of the retina, females will not be affected.

rDNA

Provide an example of gene redundancy that occurs in both eukaryotes and prokaryotes. CNV rDNA spindle fiber protein genes centromeres cell wall proteins

telomeres

Structures located at the ends of eukaryotic chromosomes are called ________. -permissive mutations -telomerases -recessive mutations -telomeres -centromeres

One light band and one intermediate band

Which of the following would result from a third round of replication using the methods of Meselson and Stahl? -One heavy band, one light band, and one intermediate band -One light band and one intermediate band -One heavy band -One light band

-The top strand (pink) is the leading strand; it is being synthesized continuously. -The bottom strand (blue) is the lagging strand; it is being synthesized discontinuously.

Q29 from chapter 11 MB* shows two antiparallel strands of DNA, replication fork moving to the left (unidirectionally) Given the direction of replication and the orientation of the two DNA strands in Part A, which of the following is true? Select the two true statements. -The top strand (pink) is the lagging strand; it is being synthesized discontinuously. -The top strand (pink) is the leading strand; it is being synthesized continuously. -The bottom strand (blue) is the lagging strand; it is being synthesized discontinuously. -The bottom strand (blue) is the leading strand; it is being synthesized continuously.

nucleoside

Regarding the structure of DNA, the covalently arranged combination of a deoxyribose and a nitrogenous base would be called a(n) ________. nucleotide monophosphate nucleoside ribonucleotide nucleoside oligonucleotide

synthesis of DNA from an RNA template

Reverse transcriptase is an enzyme found in association with retroviral activity. It has the property of ________. -synthesis of DNA from an RNA template -requiring no template -translation -most lysozymes -synthesis of RNA from a DNA template

all of the above Because eukaryotic chromosomes are linear, they possess a "telomere problem," where one daughter helix (built with the 3′→5′ lagging strand) would be shortened after each round of replication by the length of the primer. To prevent this shortening, the enzyme telomerase lengthens the 5′→3′ strand by reverse transcription, using a built-in R N A as template and guide. Now this helix's length can be maintained in the usual manner by D N A replication.

Telomerase _____. -contains its own R N A template -is a reverse transcriptase -extends one strand of the telomere -is responsible for helping to maintain chromosome size -all of the above

3' end of the parental strand

Telomerase adds nucleotides to the ________. -5' end of the parental strand -3' end of the daughter strand -3' end of the parental strand -5' end of the daughter strand

It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment.

The Hershey and Chase experiments involved the preparation of two different types of radioactively labeled phage. Which of the following best explains why two preparations were required? -The bacteriophage used in the experiments was a T2 phage. -It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. -Establishing the identity of the genetic material required observation of two phage generations. -Each scientist had his own method for labeling phage, so each conducted the same experiment using a different isotope.

XY/XX scheme

The Lygaeus mode of sex determination is the ________. -hermaphroditic scheme -XY/XX scheme -scheme based on single translocations in the X chromosome -XO/YY scheme -XX/XO scheme

After one generation in the 14N, there would be two bands, one heavy and one light (no intermediate). After the second generation in the 14N, there would also be two bands, one heavy and one light (no intermediate).

The Meselson and Stahl's experiment provided conclusive evidence for the semiconservative replication of DNA in E. coli . What pattern of bands would occur in a CsCl gradient for conservative replication? -After one generation in the 14N, there would be two bands, one heavy and one light (no intermediate). After the second generation in the 14N, there would be three bands, one heavy, one intermediate, and one light. -After one generation in the 14N, there would be one heavy band (no intermediate and no light). After the second generation in the 14N, there would be three bands, one heavy, one intermediate, and one light. -After one generation in the 14N, there would be two bands, one heavy and one light (no intermediate). After the second generation in the 14N, there would also be two bands, one heavy and one light (no intermediate). -After one generation in the 14N, there would be three bands, one heavy, one intermediate, and one light. After the second generation in the 14N, there would also be three bands, one heavy, one intermediate, and one light.

DNA helicase

The activity of ________ would be quickly undone in the absence of single-stranded binding protein. -DNA ligase -DNA polymerase I -primase -DNA polymerase III -DNA helicase

replicon

The amount of DNA that is replicated from a single origin of replication is referred to as a(n) ________. replicon ori Okazaki fragment fork primer

This nucleic acid is single-stranded DNA.

The base content of a sample of nucleic acid is as follows: A = 31%, G = 31%, T = 19%, C = 19%. What conclusion should be drawn from this information? -This nucleic acid is made of RNA. -The strands in this molecule are parallel rather than antiparallel. -This nucleic acid is single-stranded DNA. -This nucleic acid is double-stranded DNA. -The purine/pyrimidine ratio in this molecule fits Chargaff's data.

base, sugar, and phosphate

The basic structure of a nucleotide includes ________. -mRNA, rRNA, and tRNA -base, sugar, and phosphate -phosphorus and sulfur -tryptophan and leucine -amino acids

segmental deletion

The chromosomal aberration that causes cri-du-chat syndrome can be referred to as a(n) ________. -reciprocal translocation -simple translocation -segmental deletion -duplication -inversion

phosphorus and sulfur

The classic Hershey and Chase (1952) experiment that offered evidence in support of DNA being the genetic material in bacteriophages made use of which of the following labeled component(s)? phosphorus and sulfur oxygen and phosphorus tritium nitrogen and oxygen hydrogen

46, 5p-

The condition known as cri-du-chat syndrome in humans has a genetic constitution designated as ________. triploidy 46, 5p- trisomy 45, X heteroplasmy

aneuploidy

The condition that exists when an organism gains or loses one or more chromosomes but not a complete haploid set is known as ________. aneuploidy trisomy triploidy euploidy polyploidy

phosphodiester bond

The covalent linkage between the monomers in a dinucleotide is a(n) ________. ester bond disulfide bond peptide bond phosphodiester bond ionic bond

the 5′ to 3′ polarity restriction

The discontinuous aspect of replication of DNA in vivo is caused by ________. -sister-chromatid exchanges -trinucleotide repeats -polymerase slippage -the 5′ to 3′ polarity restriction -topoisomerases cutting the DNA in a random fashion

deletion or insertion During synapsis, if there is D N A found in one homolog but not in the other, a loop of the "extra" material will form. This "extra" D N A could be from the normal chromosome failing to match up to a deletion on its homolog, or the loop could form on a homolog having extra D N A (insertion) that cannot match up to the normal chromosome.

The following structure is observed during prophase 1. It could be due to which of the following conditions? pericentric inversion duplication deletion insertion deletion or insertion

B-DNA

The form of DNA that is believed to be most biologically significant is ________. D-DNA Z-DNA B-DNA E-DNA A-DNA

Z-DNA

The form of DNA that is observed to adopt a left-handed helix is ________. C-DNA A-DNA Z-DNA D-DNA P-DNA

2n = 63; euploid The horse gamete has 1n = 32 chromosomes; the donkey gamete has 1n = 31 chromosomes. When they fuse, the resulting mule cell has 32 + 31 = 63 chromosomes. Because it has complete sets of chromosomes, it is a euploid cell, even though it has an odd number of chromosomes.

The horse is 2n = 64, and the donkey is 2n = 62. If crossed, they produce the hybrid mule. There are _______chromosomes in a mule somatic cell. This cell is _________. -2n = 126; aneuploid -2n = 126; euploid -2n = 63; aneuploid -2n = 63; euploid -either 2n = 64 or 2n = 62, depending on whether the mother was the horse or the donkey; euploid.

deoxyribonucleoside triphosphate

The most specific name for an incoming monomer about to be incorporated into a DNA chain is ________. -ribonucleoside diphosphate -nucleoside monophosphate -nucleotide -deoxyribonucleoside -deoxyribonucleoside triphosphate

reverse transcription

The process of making DNA from an RNA template is referred to as ________. -recombination -reverse transcription -translation -transversion -transcription

dnaA

The protein responsible for the initial step in unwinding the DNA helix during replication of the bacterial chromosome is coded for by the ________ gene. polB polA dnaE gyrA dnaA

the fact that DNA is the genetic material

The results of the Meselson-Stahl experiments relied on all of the following except _______. -that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules -the fact that DNA is the genetic material -a means of distinguishing among the distribution patterns of newly synthesized and parent molecule DNA possible -a cesium chloride gradient

allow helicase to unwind the double helix DnaA binds to the repeats at oriC. This helps the helix to unwind which permits one helicases bind and start to break the hydrogen bonds between nitrogenous bases and denature the two D N A strands.

The role of the DnaA protein in bacterial D N A replication is to _____. -prevent renaturation of the two D N A strands during replication -synthesize an R N A primer -relieve the tension of supercoiling -allow helicase to unwind the double helix -detect replication errors

the proportion of males to females that are born

The secondary sex ratio in humans is _______. -the number of Barr bodies in a somatic cell -the proportion of males to females that are conceived -the same in all geographic areas -the proportion of males to females that are born

2:1 male to female

The sex of birds, some insects, and other organisms is determined by a ZW chromosomal arrangement in which the males have like sex chromosomes (ZZ) and females are ZW (similar to XY in humans). Assume that a recessive lethal allele on the Z chromosome causes death of an embryo in birds. What sex ratio would result in the offspring if a cross were made between a male heterozygous for the lethal allele and a normal female? 1:2 male to female 4:1 male to female 3:1 male to female 1:1 male to female 2:1 male to female

wind around each other in such a way that purines are always opposite pyrimidines and vice versa

The two DNA chains in a double helix ________. -are parallel to each other in 5' to 3' directionality -wind around each other in such a way that purines are always opposite pyrimidines and vice versa -are held together due to the charges of their phosphate groups -have an overall diameter of 2 μm -have identical base sequences

It has a single-stranded D N A genome.

The virus phi X 174 infects Escherichia coli. Its base composition is as follows: A = 24.0 % G = 23.3 % C = 21.4 % T = 31.3 % What conclusions can you draw about phi X 174 ? -It has a double-stranded D N A genome. -It has a single-stranded D N A genome. -It has a double-stranded R N A genome. -It has a single-stranded R N A genome. -A must bind to G and C to T in this virus.

D N A; purine We know this molecule is a deoxyribonucleotide, used to build D N A, because it has the sugar deoxyribose (No oxygen on Carbon 2'). The double ring is a purine ring, whereas single rings are pyrimidines.

This molecule is most likely found in ________. It has a(n) _________ ring. -R N A; purine -R N A; pyrimidine -D N A; purine -D N A; pyrimidine -either D N A or R N A; purine

One-half of the offspring would be expected to have Down syndrome.

Trisomy 21, or Down syndrome, occurs when there is a normal diploid chromosomal complement but one (extra) chromosome 21. Although fertility is reduced in both sexes, females have higher fertility rates than males. Van Dyke et al. (1995; Down Syndrome Research and Practice 3(2): 65-69) summarize data involving children born of Down syndrome individuals. Assume that children are born to a female with Down syndrome and a normal 46-chromosome male. What proportion of the offspring would be expected to have Down syndrome? -One-third of the offspring would be expected to have Down syndrome. -One-half of the offspring would be expected to have Down syndrome. -None of the offspring would be expected to have Down syndrome. -Two-thirds of the offspring would be expected to have Down syndrome. -All the children would be expected to have Down syndrome.

Two-thirds of the surviving offspring would be expected to have Down syndrome.

Trisomy 21, or Down syndrome, occurs when there is a normal diploid chromosomal complement but one (extra) chromosome 21. While there is reduced fertility in both sexes, females have higher fertility than males. Van Dyke et al. (1995; Down Syndrome Research and Practice 3(2): 65-69) summarize data involving children born of Down syndrome individuals. Given the fact that conceptuses with 48 chromosomes (four #21 chromosomes) are not likely to survive early development, what percentage of surviving offspring would be expected to have Down syndrome if both parents have Down syndrome? -Two-thirds of the surviving offspring would be expected to have Down syndrome. -One-third of the surviving offspring would be expected to have Down syndrome. -One-half of the surviving offspring would be expected to have Down syndrome. -All the children would be expected to have Down syndrome. -None of the surviving offspring would be expected to have Down syndrome.

This female would have one complete X chromosome and a Y chromosome that lacks SRY.

Under what condition might a human female have the XY sex chromosome complement? -The Y chromosome has an active Xist gene on it. -The X chromosome has a mutation such that it is able to overpower the Y chromosome. -This is not possible. -She has an XX chromosome and a Y chromosome. -This female would have one complete X chromosome and a Y chromosome that lacks SRY.

Rosalind Franklin

Watson and Crick used information from several individuals to construct their model of DNA. Whose X-ray diffraction studies were critical to their work? Erwin Chargaff Rosalind Franklin Phoebus Levene Linus Pauling

D N A pol III has 3′→5′ exonuclease activity. The ability to move "backward" over a newly synthesized strand and remove bases (exonuclease activity) that are mismatched is called proofreading. As D N A pol synthesizes D N A in the 5′→3′ direction, exonuclease activity in the 3′→5′ direction would allow proofreading.

What activity provides D N A pol III the ability to proofread? -D N A pol III has 5′→3′ exonuclease activity. -D N A pol III has 3′→5′ exonuclease activity. -D N A pol III has exonuclease activity in both the 5′→3′ and 3′→5′ directions. -D N A pol III has polymerase activity in the 5′→3′ and 3′→5′ directions. -The sliding clamp activity.

homologous regions of the X and Y chromosomes The pseudoautosomal regions refer to regions of homology between the X and Y chromosomes in humans that undergo recombination during meiosis.

What are pseudoautosomal regions on the sex chromosomes? -hemizygous regions of the X and Y chromosomes -homologous regions of the X and Y chromosomes -Regions of the X and Y chromosomes that don't recombine during meiosis -Homozygous regions of autosomal chromosomes

Some viruses contain a genetic material composed of RNA.

What are the exceptions to the general rule that DNA is the genetic material in all organisms? -Some viruses contain a genetic material composed of RNA. -Some viruses contain a genetic material composed of protein. -Some bacteria contain a genetic material composed of RNA. -Some bacteria contain a genetic material composed of protein.

RNA and protein

What are the two major components of the Tobacco Mosaic Virus? -RNA and protein -lipids and nucleic acids -RNA and DNA -carbohydrates and nucleic acids -DNA and protein

A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent.

What conclusion(s) could Griffith draw from his experiment? -A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. -The transforming factor is DNA. -The transforming factor is RNA. -The transforming factor is protein.

The Tm is the temperature at which half of the sample is denatured.

What does Tm imply? -The Tm is the temperature at which half of the sample is denatured. -The Tm is the temperature at which all the sample is denatured. -The Tm is the temperature at which the sample starts to denature. -The Tm is the maximum temperature at which the sample does not denaturate.

When the two strands in double-stranded DNA associate with each other, one strand runs 5' to 3' in one direction, the other strand runs in the same plane, but with a 3' to 5' orientation (antiparallel). Hydrogen bonding between strands occurs between adenines and thymines, and between cytosines and guanines, which are in turn called complementary base pairs.

What does it mean for a double helix of DNA to be antiparallel and complementary? -When the two strands in double-stranded DNA associate with each other, one strand runs 5' to 3' in one direction, the other strand runs in the same plane, but with a 3' to 5' orientation (antiparallel). Hydrogen bonding between strands occurs between adenines and cytosines, and between thymines and guanines, which are in turn called complementary base pairs. -When the two strands in double-stranded DNA associate with each other, one strand runs 5' to 3' in one direction, the other strand runs in the same plane, but with a 3' to 5' orientation (antiparallel). Hydrogen bonding between strands occurs between adenines and thymines, and between cytosines and guanines, which are in turn called complementary base pairs. -When the two strands in double-stranded DNA associate with each other, one strand runs 5' to 3' in one direction, the other strand runs in a different plane, but with the same (5' to 3') orientation (antiparallel). Hydrogen bonding between strands occurs between adenines and cytosines, and between thymines and guanines, which are in turn called complementary base pairs. -When the two strands in double-stranded DNA associate with each other, one strand runs 5' to 3' in one direction, the other strand runs in a different plane, but with the same (5' to 3') orientation (antiparallel). Hydrogen bonding between strands occurs between adenines and thymines, and between cytosines and guanines, which are in turn called complementary base pairs.

B and C: SRY and the centromere

What does the nonrecombining region of the Y contain? -the genes that have homologues on the X chromosome -SRY -the centromere -A, B, and C -B and C

ZZ is a homogametic male and ZW is a heterogametic female.

What does the notation ZZ/ZW indicate regarding sex determination? -Z corresponds to the human X chromosome and W to the Y chromosome. -ZZ is a homogametic male and ZW is a heterogametic female. -ZZ is a hermaphrodite and ZW is a male. -ZZ is a homogametic female and ZW is a heterogametic male.

that length of D N A that is replicated by the core enzyme before it detaches from the template The sliding D N A clamp component of D N A polymerase III allows the core enzyme to replicate much longer pieces of D N A than would occur in its absence.

What does the term "processive" mean in the context of D N A replication? -that the D N A polymerase III holoenzyme has completed its self-assembly -that the D N A polymerase III has completed one round of D N A replication -that the D N A polymerase III is replicating bidirectionally -that length of D N A that is replicated by the core enzyme before it detaches from the template -the ability to move "backward" for proofreading.

In some species, more males will be produced when the eggs are incubated at higher temperatures, while in others a greater number of females will result.

What effect does the rise in temperature, that is associated with global warming, likely to have on sex ratios of reptiles? -No difference in the proportion of females to males will result as only colder temperatures have been shown to affect sex determination. -More females will be produced. -More males will result. -In some species, more males will be produced when the eggs are incubated at higher temperatures, while in others a greater number of females will result.

D N A pol I While D N A pol III performs the bulk of D N A synthesis, D N A pol I has the special 5′→3′ exonuclease activity which allows it to remove R N A primers

What enzyme will replace the R N A primers found in the newly synthesized strand? D N A pol III D N A pol II D N A pol I Primase ligase

unequal crossing over

What error of meiosis leads to both a duplication and a deletion? -replication cross formation -unequal crossing over -X-ray chromosomal breakage -D loop formation -replication errors

Monosomy may unmark recessive lethals that are tolerated in heterozygotes carrying the wild-type allele.

What explanation is generally given for lethality of monosomic individuals? -The loss of a single chromosome is not generally lethal, unless the individual is inbred. -The gametes of monosomic individuals cannot undergo meiosis, and this is lethal. -Monosomy may unmark recessive lethals that are tolerated in heterozygotes carrying the wild-type allele. -Monosomic chromosome cannot undergo mitosis correctly. -Cells count the number of chromosomes they have and will undergo apoptosis when the chromosome number is incorrect.

a structural component of a chromosome due to the nucleotide sequence to it that leads to a high percentage of chromosomal breakage

What is a fragile site? -the telomere of a chromosome can fall off due to spindle malfunctions during meiosis -a structural component of a chromosome due to the nucleotide sequence to it that leads to a high percentage of chromosomal breakage -an area around the centromere prone to inversion due to gene duplications -a structural component of a chromosome due to proteins bound to it that leads to a high percentage of chromosomal breakage -a site in a chromosome prone to duplication, such as the bar gene

The designed loss of a specific gene in order to study its function.

What is meant by a gene knockout? -The synthesis of gene DNA using RNA as a template. -The designed loss of a specific gene in order to study its function. -The deletion of a gene in order to study the structure of the encoded protein. -The insertion of several six-nucleotide repeats in the gene sequence in order to study the structure of the encoded protein.

The strength of the association between the enzyme and its substrate, and thus the length of DNA that is synthesized before the enzyme dissociates from the template.

What is meant by processivity of a DNA-synthesizing enzyme? -The strength of the association between the enzyme and its substrate, and thus the length of DNA that is synthesized before the enzyme dissociates from the template. -The strength of the association between the enzyme and DNA, and thus the number of new DNA molecules that is synthesized before the enzyme dissociates from the template. -The strength of the association between the enzyme and dNTPs, and thus the speed of the new DNA strand synthesis. -The strength of the association between the enzyme and its cofactors, and thus the speed of the new DNA strand synthesis.

A hyperchromic effect is the increased absorption of UV light as double-stranded DNA is converted to single-stranded DNA.

What is the hyperchromic effect? -A hyperchromic effect is the decreased absorption of UV light as single-stranded DNA is converted to double-stranded DNA. -A hyperchromic effect is the decreased absorption of UV light as double-stranded DNA is converted to single-stranded DNA. -A hyperchromic effect is the increased absorption of UV light as double-stranded DNA is converted to single-stranded DNA. -A hyperchromic effect is the increased absorption of UV light as single-stranded DNA is converted to double-stranded DNA.

The nondisjunctional event that produces Down syndrome occurs more frequently during oogenesis in women older than age 35.

What is the leading cause of Down syndrome? -Aberrant implantation in the uterus leads to developmental problems in the fetus. -The nondisjunctional event that produces Down syndrome occurs more frequently during oogenesis in women older than age 35. -In men over 25, sperm formation is impaired and produces monosomic children. -In older parents, there is egg/sperm incompatibility leading to duplication of chromosome 21. -In older females, chromosome 21 is duplicated leading to abnormal egg formation.

single-stranded

What is the physical state of DNA after it is heated and denatured? -individual nucleotides -double-stranded -individual bases, sugars, and phosphates -single-stranded

primary sex ratio

What is the ratio of male to female conceptions in a population called? -anomalous gender reorder calculation -conception birth order number -primary sex ratio -secondary sex ratio -gender order

It converts androgens to estrogens.

What is the role of the enzyme aromatase in sexual differentiation in reptiles? -It converts estrogens to androgens. -It destroys excess estrogens in males. -It regulates translation of androgen genes. -It converts androgens to estrogens. -It regulates transcription of estrogen genes.

D N A polymerase III Replication is performed by D N A polymerases. This must be Pol III, as we see synthesis of both leading and lagging strands.

What protein is functioning at the point marked with the large arrow? (large green broken sphere on the lagging strand) a.D N A ligase b.helicase c.gyrase d.D N A polymerase I e.D N A polymerase III

Xic

What region is the major control center of X inactivation in mammals? NRY Barr Body SRY MSY Xic

DNase destroyed the transforming activity.

What results did Avery, McLeod, and McCarty obtain in their experiments with virulent bacteria? -Protease destroyed the transforming activity. -RNase destroyed the transforming activity. -The transforming principle was too complex and difficult to be purified. -DNase destroyed the transforming activity.

inversion (paracentric) heterozygote

What type of chromosomal configuration does the following diagram illustrate? (#17 on Canvas review questions) -simple translocation heterozygote -inversion (paricentric) heterozygote -inversion (paracentric) heterozygote -reciprocal translocation heterozygote -inversion (paracentric) homozygote

It was instrumental in demonstrating that DNA polymerase I activity is not necessary for in vivo replication.

What was the significance of the polA1 mutation? -It was instrumental in demonstrating that DNA polymerase I activity is not necessary for in vivo replication. -It was instrumental in estimating the rate of DNA synthesis in bacteria. -It was used to determine the functions of different subunits of DNA polymerase I. -It was used to pinpoint the region of the chromosome that contains the gene determining the rate of DNA synthesis in bacteria.

to determine the exact molecular species of the "transforming principle"

When Avery and his colleagues had obtained what was concluded to be the transforming factor from the IIIS virulent cells, they treated the fraction with proteases, RNaseRNase, and DNaseDNase, followed by the assay for retention or loss of transforming ability. What was the purpose of these experiments? -to prove the stability of the "transforming priniciple" in a typical cell medium -to prove that transforming ability does not depend on presence of any enzymes -to determine the enzyme that can suppress virulence -to determine the exact molecular species of the "transforming principle"

cells lining the gut of mosquito larvae attain 16n ploidy

Which of the following is an example of an endopolyploid? -two diploid mate and produce a tetraploid -an egg has complete nondisjunction and the resultant offspring is triploid -cells lining the gut of mosquito larvae attain 16n ploidy -an individual suffers from a trisomy -one chromosome is lost during cell division

The DNA was proven to be the "transforming principle". Treatment with DNase resulted in the loss of transforming ability.

When Avery and his colleagues had obtained what was concluded to be the transforming factor from the IIIS virulent cells, they treated the fraction with proteases, RNaseRNase, and DNaseDNase, followed by the assay for retention or loss of transforming ability. What was the result? What conclusions were drawn? -The RNA was proven to be the "transforming principle". Treatment with RNase resulted in the loss of transforming ability. -The DNase was proven to be the "transforming principle". Treatment with DNase resulted in the retention of transforming ability. -Protein was proven to be the "transforming principle". Treatment with protease resulted in the loss of transforming ability. -The DNA was proven to be the "transforming principle". Treatment with DNase resulted in the loss of transforming ability.

transfection

When a protoplast is infected by only the nucleic acid component of a virus, the infection is referred to as ________. recombination transduction transformation transfection lysis

attached to the nuclear envelope

Where is the general location of a Barr body? -attached to the other X chromosome -attached to the nuclear envelope -in the nucleolus -stuck in a nuclear pore -adjacent to the Y chromosome

DNA polymerase III

Which DNA polymerase is mainly responsible for genome replication in E. coli? DNA polymerase III DNA polymerase α DNA polymerase II DNA polymerase I

3' to 5'; exonuclease

Which activity of E. coli DNA polymerase I is responsible for proofreading the newly synthesized DNA? 3' to 5'; exonuclease 3' to 5' endonuclease 5' to 3' exonuclease 5' to 3' polymerase

fixed point of initiation, bidirectional, semiconservative

Which cluster of terms accurately reflects the nature of DNA replication in prokaryotes? -random point of initiation, unidirectional, semiconservative -random point of initiation, bidirectional, semiconservative -fixed point of initiation, unidirectional, conservative -fixed point of initiation, bidirectional, semiconservative -fixed point of initiation, bidirectional, conservative

DNA polymerase

Which enzyme catalyzes the addition of nucleotides to a growing DNA chain? Primase DNA polymerase Telomerase Helicase

Homologs fail to separate during meiosis I.

Which of the following accurately describes a possible meiotic nondisjunction event? -Sister chromatids fail to separate during meiosis I. -Homologs fail to separate during meiosis I. -Fusion of gametes results in trisomy. -Meiosis fails to proceed to completion.

One of the two female X chromosomes is randomly inactivated in an early embryonic stage.

Which of the following best describes the Lyon hypothesis? a.Barr bodies are inactivated X chromosomes. b.The paternal X chromosome of a female is inactivated in an early embryonic stage. c.The maternal X chromosome of a female is inactivated in an early embryonic stage. d.One of the two female X chromosomes is randomly inactivated in an early embryonic stage. e.Each cell of a female randomly inactivates one X chromosome shortly after mitosis.

-Deletion

Which of the following causes cri-du-chat syndrome? -Deletion -Duplication -Insertion -Nonreciprocal translocation -Reciprocal translocation

meiosis

Which of the following characteristics is common to species that reproduce sexually? a.morphologically distinguishable male and female gametes b.diploid adult stage c.different sets of sex chromosomes in males and females d.Y chromosome in males e.meiosis

double-stranded, antiparallel, (A + T)/C + G) = variable, (A + G)/(C+ T) = 1.0

Which of the following clusters of terms accurately describes DNA as it is generally viewed to exist in prokaryotes and eukaryotes? -double-stranded, antiparallel, (A + T)/C + G) = variable, (A + G)/(C+ T) = 1.0 -single-stranded, antiparallel, (A + T)/C + G) = 1.0, (A + G)/(C + T) = 1.0 -double-stranded, antiparallel, (A + T)/C + G) = variable, (A + G)/(C + T) = variable -double-stranded, parallel, (A + T)/C + G) = 1.0, (A + G)/(C + T) = 1.0 -double-stranded, parallel, (A + T)/C + G) = variable, (A + G)/(C + T) = 1.0

Familial Down syndrome

Which of the following conditions results from a Robertsonian translocation? -Familial Down syndrome -The Bar mutation in Drosophila -Klinefelter syndrome -Cri du chat syndrome

Turner syndrome

Which of the following human genetic conditions is missing a chromosome? Jacob syndrome Klinefelter syndrome XXXX syndrome Down's syndrome Turner syndrome

It is composed of protein.

Which of the following is NOT a characteristic of the genetic material? -It contains all the information needed for growth, development, and reproduction of the organism. -It must be replicated accurately. -It must be capable of change. -It is composed of protein.

Both typically result from nondisjunction. The genotype associated with Klinefelter syndrome is 47, XXY, while that of Turner syndrome is 45, X.

Which of the following is TRUE of both Klinefelter syndrome and Turner syndrome? -Both typically result from nondisjunction. -It is the mechanism by which dizygotic, but not monozygotic, twins are produced. -Both give rise to individuals who have external male genitalia. -Both are the result of mitotic errors early in development.

they may lead to translocation cross formation during synapsis

Which of the following is not a potential outcome of a gene duplication? -they may result in providing the raw material for evolution -they may lead to translocation cross formation during synapsis -they may lead to the development of gene families -they may produce phenotypic variation -they may result in gene redundancy

X inactivation occurs on the maternal X chromosome

Which of the following is not a tenant of the Lyon hypothesis? -X inactivation is passed down during mitosis. -X inactivation occurs early in development. -X inactivation of descendant cells is the same as the progenitor cell. -initial X inactivation is random. -X inactivation occurs on the maternal X chromosome.

(C + G) = (A + T)

Which of the following is not consistent with Erwin Chargaff's findings? -(A + G) = (C + T) -(A + C) = (G + T) -A = T -(C + G) = (A + T) -(A + G)/(C + T) = 1

The strands run parallel

Which of the following is not included in the Watson-Crick model of D N A structure? -It is composed of two strands. -The two strands are held together by H bonds between nitrogenous bases. -The strands run parallel. -The resultant helix is right-handed. -The helix has a constant diameter of 20 Å.

XXY and XYY

Which of the following karyotypes would lead to male characteristics in humans? XYY XXY XO XO and XYY XXY and XYY

Both preparations of infected bacteria would exhibit radioactivity.

Which of the following outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender? -Both preparations of infected bacteria would contain both P32 and S35. -The phage would fail to infect bacteria. -Neither preparation of infected bacteria would exhibit radioactivity. -Both preparations of infected bacteria would exhibit radioactivity.

Single‑strand binding proteins stabilize the open conformation of the unwound DNA.

Which of the following statements about DNA replication is true? -DNA gyrase unwinds the DNA double helix. -Okazaki fragments are DNA fragments synthesized on the leading strand. -Single‑strand binding proteins stabilize the open conformation of the unwound DNA. -DNA polymerase adds dNTP monomers in the 3′-5′ direction.

All of the above are false. In most cases, individuals with Down syndrome have 47 chromosomes, with three copies of chromosome 21, or trisomy 21. This could occur due to nondisjunction in either parent. A triploid person (which is a lethal condition in humans) would have three complete sets of chromosomes, or 69 chromosomes. While the incidence of Down syndrome births increases in women over 35, most such births occur in younger women. Familial Down syndrome is due to translocation of chromosome 21 onto chromosome 14.

Which of the following statements about Down syndrome is false? -Individuals with Down syndrome are triploid. -Down syndrome occurs due to nondisjunction in either meiotic division in the mother only. -Most Down syndrome births happen to women over age 35. -Familial Down syndrome is due to nondisjunction. -All of the above are false.

They are generally taller than average.

Which of the following statements about XYY males is true? -They are generally taller than average. -They are sterile. -They have underdeveloped secondary sex characteristics. -They are likely to end up in prison.

They may be sterile and unable to produce offspring.

Which of the following statements about allopolyploid individuals is true? -They result from a mating between individuals of the same species. -They are generated from exposure to colchicine. -They can be formed when two sperm simultaneously fertilize an ovum within the same species. -They may be sterile and unable to produce offspring.

Normally in humans, all the sons of a male showing an X-linked phenotype will inherit the trait.

Which of the following statements below is false -Normally in humans, all the sons of a male showing an X-linked phenotype will inherit the trait. -At meiosis I, the X and Y chromosomes line up as if they were homologs. -Normally in humans, females are carriers of X-linked recessive traits if they are heterozygous. -Normally in humans, all the sons of a female homozygous for an X-linked recessive gene will inherit that trait. -Normally in humans, a male that is carrying an X-linked dominant trait will pass it to all his daughters.

DNA is the identity of the hereditary material in phage T2.

Which of the following statements best represents the central conclusion of the Hershey-Chase experiments? -When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates radioactive label. -Some viruses can infect bacteria. -Phage T2 is capable of replicating within a bacterial host. -DNA is the identity of the hereditary material in phage T2.

Familial Down syndrome is caused by a translocation involving chromosome 21.

Which of the following statements is TRUE? -Individuals with familial Down syndrome are trisomic and have 47 chromosomes. -Inversions and translocations are without evolutionary significance. -In general, inversion and translocation heterozygotes are as fertile as organisms whose chromosomes are in the standard arrangement. -Translocations may be pericentric or paracentric. -Familial Down syndrome is caused by a translocation involving chromosome 21.

Dosage compensation is accomplished in humans by inactivation of the Y chromosome.

Which of the following statements is false? -An individual with Turner Syndrome has no Barr bodies. -Dosage compensation is accomplished in humans by inactivation of the Y chromosome. -A typical XX human female has one Barr body. -An individual with Klinefelter syndrome generally has one Barr body. -In a cell with X chromosomes lacking the XIC, there is still X inactivation.

They are short fragments of DNA synthesized from RNA primers on the lagging strand. Okazaki fragments are short sequences synthesized in the lagging strand because DNA polymerase can synthesize only from 5' to 3' and the DNA strands are antiparallel.

Which of the following statements is true regarding Okazaki fragments? -They are synthesized by ligase. -They are short fragments of RNA on the leading strand. -They are short fragments of DNA synthesized from RNA primers on the lagging strand. -They add nucleotides to the elongating DNA.

Trisomic and aneuploid

Which of the following terms can be used to describe Down syndrome? Triploid and polyploid Trisomic and aneuploid Trisomic and polyploid Triploid and aneuploid

Avery, MacLeod, and McCarty

Who discovered that DNA was the genetic material (or "transforming principle") that could convert nonvirulent R-type Diplococcus pneumoniae bacterium to the virulent S-type? -Watson, Crick, Wilkins, and Franklin -Frederick Griffith -Erwin Chargaff -Avery, MacLeod, and McCarty -Hershey and Chase

32P labeled DNA only; 35S labeled protein only.

Why did Hershey and Chase use the isotopes 32P and 35S in their experiments? -35S labeled DNA only; 32P labeled protein only. -32P labeled DNA only; 35S labeled protein only. -32P labeled DNA only; 35S labeled carbohydrates only. -35S labeled lipids only; 32P labeled DNA only.

G-C base pairs have three hydrogen bonds while A-T base pairs have two hydrogen bonds.

Why is Tm related to base composition? Which statement is true? -G-C base pairs have three hydrogen bonds while A-T base pairs have two hydrogen bonds. -A-T base pairs have three hydrogen bonds while G-C base pairs have two hydrogen bonds.

The more hydrogen bonds between bases, the higher the temperature needed to separate the pair.

Why is Tm related to base composition? Which statement is true? -The fewer hydrogen bonds between bases, the higher the temperature needed to separate the pair. -The more hydrogen bonds between bases, the higher the temperature needed to separate the pair.

phenotypic mosaicism

X-linked genes in female mammals often demonstrate ________. -underrepresentation in the genome -incomplete dominance -nonrecombination during meiosis -phenotypic mosaicism -heterochromatin formation

The mutant chromosome would always be active.

XistRNAplays a crucial role in X inactivation. The Xist RNA coats the X chromosome that produced it, and whichever X has more Xist, it becomes the inactivate X. If a mutation inactivated Xist on only one chromosome, what would you predict to occur? a.The mutant chromosome would always be inactivated. b.The mutant chromosome would always be active. c.Both chromosomes would be inactivated. d.Both chromosomes would be active. e.The wild type Xist would compensate for the mutant, and X inactivation would proceed normally.

Males transcribe their X at twice the rate of a female X.

You observe the cells and karyotypes of a novel species of mammal found on an island. No females have a Barr body; however, they all have two X chromosomes. Males have one X and one Y chromosome. Considering dosage compensation, which of the following would you predict? a.Females only transcribe one X. b.Males transcribe their X at twice the rate of a female X. c.Males transcribe their X at half the rate of a female X. d.Females transcribe their X at twice the rate of a male X. e.Transcription rates would not be affected.

Nondisjunction

________ is viewed as a major cause of aneuploidy. Colchicine treatment Nondisjunction Heat treatment Segmental deletions X-ray mutations