The mouse survived injection of live virulent (smooth) Streptococcus pneumoniae.
That DNA is the genetic material.
The mouse did not survive when injected with a mixture of live, avirulent (smooth) Streptococcus pneumoniae and heat-killed virulent Streptococcus pneumoniae.
The heat-killed, virulent Streptococcus pneumoniae was lethal to the mouse.
Protease destroyed the transforming activity.
DNase destroyed the transforming activity.
RNase destroyed the transforming activity.
The transforming principle was too complex and difficult to be purified.
Thymine can never be attached to ribose.
Thymine is the same as 5-methyl uracil.
Thymine has an amino group at the C-4 position.
Thymine has a methyl group attached to the N-1 position.
The amount of the other purine adenine must also equal 20%.
The amount of cytosine must be equal to 30% so that the total amount of G + C equals 50%.
The amount of thymine must equal 30%.
The ratio of purines to pyrimiidines must be greater than 1.
The sugar phosphate backbone of Z-DNA assumes a zigzag course through space.
Z-DNA is a typical of GC-rich sequences.
It forms a left-handed helix.
The strands are not antiparallel.
The sugar molecule.
The 5'-3' orientation of the polynucleotide strand.
The presence of uracil.
The number of different functions performed.
X-ray diffraction analysis
A longer DNA molecule would associate more quickly (smaller Cot1/2 value).
Molecules with the highest GC content would associate more slowly.
DNA with more repeated sequences requires more time to reassociate (smaller Cot value).
Shorter DNA molecules would associate more quickly (smaller Cot1/2 values).
Heat-killed cultures treated with DNase would transform the R cells.
Heat-killed cultures treated with protease would not transform the R cells.
Heat-killed cultures treated with RNase would transform the R cells.
Heat-killed cultures treated with protease would transform the R cells.
No conclusion is possible.
DNA and protein together serve as the genetic material.
DNA is the genetic material.
Protein is the genetic material.
C-3' and C-5'
C-1', C-2', and C-3'
C-1' and C-5'
DNA is composed of a sugar-phosphate backbone with bases projecting toward the inside of the backbone.
DNA is composed of a sugar-phosphate backbone with bases projecting toward the outside of the backbone.
DNA is composed of a sugar-phosphate backbone made up of bases hydrogen-bonded to each other.
DNA is composed of a deoxynucleoside triphosphate with a base attached to it.
DNA with a low AT content would melt more slowly.
All DNA strands of equal length have equal melting temperatures.
The DNA with the greater number of repetitive sequences will melt more slowly (lower Tm).
DNA with a high GC content must have a lower Tm.
RNA is usually much longer than DNA.
DNA is usually double-stranded.
RNA has a free 2' hydroxyl group.
The DNA base composition must have the amount of purines equal to the amount of pyrimidines.
It is not important; the virus can make its own reverse transcriptase from its viral RNA after infection.
The viral reverse transcriptase is different from the host cell's reverse transcriptase.
The retrovirus must make DNA before it can use the cell's molecular machinery to make reverse transcriptase. Normal cells do not contain reverse transcriptase.
The pentose sugar in DNA is ribose.
Nucleic acids are formed through phosphodiester bonds that link nucleosides together.
Hydrogen bonds formed between the sugar-phosphate backbone of the two DNA chains help to stabilize DNA structure.
The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.
5' TACGAACTGAC 3'
5' CAGTCAAGCAT 3'
5' ATGCTTGACTG 3'
5' ACTCTACGTAG 3'
All white‑leafed progeny
White‑ and green‑leafed progeny
All variegated‑leafed progeny
All green‑leafed progeny
All white‑leafed progeny
White‑ and green‑leafed progeny
All variegated‑leafed progeny
All green‑leafed progeny
Mitochondria and chloroplasts arose independently from free‑living bacteria.
The ancestral bacteria from which mitochondria and chloroplasts are derived were capable of photosynthesis and aerobic respiration.
The entire genome of the ancestral bacteria has been preserved in present‑day mitochondria and chloroplasts.
Eukaryotic cells engulfed the ancestral bacteria from which mitochondria and chloroplasts are derived, and the two entities formed a symbiotic relationship.
Leber's hereditary optic neuropathy
Petite mutations in Saccharomyces cerevisiae
Shell coiling in Limnaea peregra
Variegation in Mirabilis jalapa
High female sex ratio in Drosophila bifasciata and Drosophila willistoni
It is a sex-influenced trait.
The trait is an example of maternal inheritance.
It is an X-linked trait.
The trait is an example of maternal influence.
Left‑hand coiled Dd female × left‑hand coiled dd male
Left‑hand coiled DD female × left‑hand coiled dd male
Right‑hand coiled dd female × left‑hand coiled DD male
Left‑hand coiled DD female × right‑hand coiled dd male
Left‑hand coiled DD female × right‑hand coiled DD male
Right‑hand coiled dd female × right‑hand coiled dd male
Left‑hand coiled dd female × right‑hand coiled DD male
Right‑hand coiled Dd female × right‑hand coiled Dd male
Petite in Saccharomyces cerevisiae
Kearns–Sayre syndrome in humans
Poky in Neurospora crassa
Streptomycin resistance in Chlamydomonas reinhardi
The mother was heterozygous, and the gamete that formed the zygote did not carry the mutant allele.
The mother exhibited incomplete penetrance for the trait.
The disease does not affect females, only males.
A new mutation arose in the mitochondrial genome.
All streptomycin‑sensitive mt+
All streptomycin‑resistant mt–
One‑half streptomycin‑resistant mt+ and one‑half streptomycin‑resistant mt–
One‑half streptomycin‑sensitive mt+and one‑half streptomycin‑sensitive mt–
In maternal inheritance, the phenotype of the progeny does not necessarily depend on the individual's genotype but on the genotype of the maternal parent.
In maternal inheritance, reciprocal crosses give identical results but with different ratios among male and female progeny.
In maternal inheritance, the genes are linked to the X chromosome.
In maternal inheritance, a trait is transmitted through the ooplasm, because the mitochondria and chloroplasts of the zygote derive from the ooplasm.
All haploid gametes
Two haploid gametes, one diploid gamete, and one gamete missing a copy of one of the chromosomes
Two diploid gametes and two haploid gametes
All trisomic gametes
Chromosomes 21, 18, and 13
Only the sex chromosomes X and Y
X, Y, and chromosome 21
Fragile X syndrome
Inclusion of the centromere in the inversion.
One suppresses crossovers; the other encourages crossovers.
The placement of the centromere in the inversion.
The formation of dicentric bridges.
Down syndrome resulting from a Robertsonian translocation does not result from nondisjunction, which is affected by age.
This statement is incorrect; the age of the mother does affect the transmission of Down syndrome resulting from Robertsonian translocations.
Mothers with a Robertsonian translocation never have children.
Robertsonian translocations occur only in young mothers.
It separates maternal from paternal chromatids.
It occurs more frequently in human females over age 35.
It may fail to separate maternal chromatids from one another or paternal chromatids from one another.
It may fail to separate maternal from paternal chromatids.
Familial Down syndrome
Fragile X syndrome
Triploid and aneuploid
Trisomy and polyploid
Trisomy and aneuploid
Triploid and polyploid
A chromosome break and exchange
Uneven crossing over during meiotic prophase
Spindle failure during chromosome separation
Loss of a telomere
The large chromosome represents a deleted chromosome number 1.
The large chromosome represents a Robertsonian fusion between chromosomes 21 and 18.
There is too little information about chromosome 21 to draw a conclusion.
The large chromosome represents an extra chromosome 3.
Nondisjunction during either meiosis I or II in the male gamete
Nondisjunction during meiosis I in either the male and female gamete
Nondisjunction during meiosis II in either the male and female gamete
Nondisjunction during either meiosis I or II in the female gamete
Diploid number of 5
Diploid number of 12
Haploid number of 5
Haploid number of 10
Normal, even though they have too much genetic material
Abnormal, because they have too much genetic material
Normal, because they have a normal amount of genetic material
Abnormal, because they lack some genetic material
Fragile X syndrome-deletion
Down syndrome-Robertsonian translocation
The two modes of reproduction maximize the genetic variation within a species.
The organisms are hermaphroditic.
The two modes of reproduction are favored under different conditions.
The organisms are monoecious.
Occurs only within the megaspore mother cell.
Is diploid and produces haploid spores.
Is linked to the haploid phase via mitosis.
Is diploid and produces diploid spores.
No individuals are exclusively female.
The lineage of every adult cell can be traced back to its embryonic origins.
Adults are very simple, consisting of approximately 1000 cells.
Males have a Y chromosome, and hermaphrodites have an X chromosome.
They are generally taller than average.
They are sterile.
They are likely to end up in prison.
They have underdeveloped secondary sex characteristics.
By containing the autosomal genes responsible for the SRY effects
By containing the genes for steroid metabolism
By containing a gene coding for the TDF, which directs the development of fetal gonads
By containing the structural genes for testes production
Individuals with either syndrome have 44 autosomes.
Both syndromes can be identified by a karyotype.
Individuals with either syndrome are both identified as females.
Both syndromes result from nondisjunction.
Regions of the X and Y chromosomes that don't recombine during meiosis
Homologous regions of the X and Y chromosomes
Homozygous regions of autosomal chromosomes
Hemizygous regions of the X and Y chromosomes
Women live significantly longer than men.
Y-bearing sperm are better swimmers than X-bearing sperm.
More boys die during infancy than girls.
There is disproportionate death of female fetuses prior to birth.
They contain no DNA.
They are inactivated Y chromosomes.
They ensure that males and females have equal "doses" of the genes on the X chromosome.
They are visible only in males.
By the ratio of the number of X chromosomes to the number of sets of autosomes
By the ratio of the number of Y chromosomes to the number of sets of autosomes
By the ratio of the number of X chromosomes to the number of autosomes
By the ratio of the number of Y chromosomes to the number of autosomes
Male or female.
Produced via mitosis.
It results in individuals with an extra copy of one of the sex chromosomes.
It is the mechanism by which monozygotic, but not dizygotic, twins are produced.
It is the mechanism by which two sperm nuclei fertilize an oocyte and two endosperm nuclei, respectively.
It is the mechanism by which dizygotic, but not monozygotic, twins are produced.
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 temperature at which the eggs are incubated.
The number of Barr bodies present.
Females are mosaics for homozygous X-linked genes.
Females are mosaics for heterozygous X-linked genes.
Males are mosaics for hemizygous X-linked genes.
Females are mosaics for heterozygous autosomal genes.
It is a problem in species that have more autosomes than sex chromosomes.
It cannot be directed by a master switch that enhances X-chromosome activity.
Its mechanism is the same in all animals.
It must occur when one sex has more copies of a gene or genes than the other sex.
They can be induced via exposure to the bacteriophage T1.
They are expressed directly in descendant cells because bacteria are haploid.
They occur at a much higher frequency.
They are visible as color changes within a bacterial plaque.
All of these choices
Nutrients and oxygen
They function as the donor in crosses with F− bacteria.
They are derived from F+ bacteria.
They frequently convert recipient bacteria to F+ cells.
They function as the donor only in crosses with F− bacteria and are so called because they result in high frequency recombination.
A haploid bacterial cell
The result of sexual reproduction between two bacteria cells
A partially diploid bacterial cell
A zygote formed from a bacteriophage and a bacterial cell
Unlinked genes are passed simultaneously from one cell to another.
Two bacterial cells within a culture are transformed by the same genetic material.
A bacterial cell receives two adjacent genes on a single piece of DNA from the medium.
It is not uncommon for the entire bacterial chromosome and F factor to be passed from one cell to another.
They are viruses.
They can pick up bacterial genes during the process of infecting a bacterial cell.
Within their heads, they can pack the main chromosome from one cell and transfer it to another.
They use the metabolic machinery of a bacterial cell to produce more copies of themselves.
An infected bacterial cell is immediately ruptured following rapid replication of the phage DNA.
A phage infects a bacterium, and the phage DNA lives symbiotically within the F‑factor plasmid in the cell.
Viral DNA is integrated into the bacterial DNA and passed on harmlessly with each round of replication and cell division.
Chemical or UV light treatment renders the viral DNA harmless to the bacterial cell.
Viral‑mediated viral recombination
Bacterial‑mediated bacterial recombination
Viral‑mediated bacterial recombination
Bacterial‑mediated viral recombination
Mapping genes by observing recombination between a virus and a bacterial genome
Gene mapping within a species
Mapping between genes
Mapping genes by observing recombination between two different virus genomes
Sister chromatids, not homologous chromosomes, exchange information during recombination.
A portion of DNA from one genome is replaced with homologous DNA from another strain of bacteria.
Only the plasmid DNA goes through recombination in prokaryotes.
Recombination does not occur in prokaryotes.
Bidirectionally between two cells
From one bacterium to another
With the help of a viral go‑between
From one cell into the culture medium, where it is taken up by another cell
The original gene, still floating in the cytoplasm, may be used when replicating the chromosome.
At transformation, a heteroduplex is formed that alters only one strand of the DNA, leaving the other unchanged.
Many cells reject newly integrated genes.
Transformed cells are initially heterozygous.
The F factor is removed from the chromosome, taking chromosomal genes with it, and the cell then transfers its F plasmid to another cell.
They are simultaneously infected by two Hfr strains.
None of these choices. They are always haploid.
They are infected with a transforming phage that has been incubated with bacteria from the same strain.
They do not require conjugation for gene transfer.
Rather than existing as an isolated plasmid, the F factor is incorporated into the bacterial chromosome.
They contain two distinct F factors.
They induce the production of significantly more sex pili than normal F+ cells.
F−cells become F+ or Hfr cells only if the mating is interrupted before a complete chromosome is transferred.
Some Hfr cells do not carry the F factor.
Hfr cells often carry the F factor in a free‑floating plasmid that is not transferred.
The F factor is the last part of the chromosome to be transferred.
48 × 10^7
4.8 × 10^−7
48 × 10^−6
4.8 × 10^6
DNA is transferred from an F+ cell to an F− cell.
Only competent cells can undergo conjugation.
The F factor is an element that is found in the chromosome of an F+ cell.
One strand of DNA from an F+ cell integrates into the chromosome of an F− cell, and the other strand is degraded.
The F factor is inserted into the chromosome of an F− cell, causing it to become an Hfr strain.
The F factor is excised from the chromosome of an Hfr strain, causing it to revert to F−.
The F factor and several adjacent genes are excised from the chromosome of an F+ cell and transferred to an F− strain.
The F factor is inserted into the chromosome of an F+ cell, causing it to revert to F−.
The phage produces lysozyme, which ruptures the host cell wall and releases newly formed viral particles.
The bacterial DNA is degraded.
The phage takes over the protein synthesis machinery of the bacteria.
The phage DNA is injected into the host and integrates into the bacterial chromosome.
The plate would have a few plaques.
The entire lawn of bacteria would be lysed.
None of these choices
The plate would have no plaques.