Genetics

c) If crossed with another heterozygous plant, the majority of progeny will have the dominant flower color. There will be roughly three times as many plants with the dominant flower color as plants with the recessive flower color.

True The 1:2:1 genotypic ratio represents relative probabilities of gamete combinations based on the assumption that gamete union is random.

a) Observe the relevant phenotype in the progeny that result from a cross between individuals from two different pure-breeding lines. All progeny will be heterozygous for the trait in question and will display the phenotype that corresponds with the dominant allele.

True The probability of having a boy is ½, and the probability of having a girl is ½. The sex of the previous children does not influence the sex of the third child.

c) 20/64 The term of interest in the expansion (a + b)^6 is the middle term: (a^3)(b^3). Using Pascal's triangle, the coefficient of this term is 20, so the probability of three boys and three girls is 20⁄64.

d) 1/216 Each die is a different color, and thus the probability of rolling a 6 on all three is the product of the individual probabilities: 1⁄6 × 1⁄6 × 1⁄6.

c) The phenotypic ratio among phenotypes produced from an F1 X F1 dihybrid cross is 9:3:3:1. The 9:3:3:1 ratio requires that the four gamete classes produced occur with equal frequency. This requires independent assortment.

False

a) 1/4 There are four different possible genotypes. Independent Assortment assumes that all are equally likely. Therefore, each of the four should occur with roughly equal frequency (1/4).

True Only the dominant allele is expressed in an individual that is heterozygous for a particular trait.

c) 3 round seeds, 1 wrinkled seed The F1 generation would have the genotype Rr, so crossing two heterozygotes would result in 3 plants with round seeds and 1 plant with wrinkled seeds.

d) 25% The proportion of rr is ¼, so the next offspring has a 25% chance of being wrinkled.

True The heterozygous offspring exhibit a phenotype that is intermediate between the parents.

c) The individual lacks the enzyme required to produce the H substance. Individuals with the Bombay phenotype are homozygous hh and do not make the enzyme required to produce the H substance. Thus, they cannot modify the H substance to produce the antigens on which type A and B blood phenotypes are based.

c) 4⁄16 disc, 8⁄16 sphere, 4⁄16 long The genotypic ratio for this cross is 4⁄16 AB, 4⁄16 Ab, 4⁄16 aB, and 4⁄16 ab, making the phenotypic ratio 4⁄16 disc, 8⁄16 sphere, 4⁄16 long.

c) The differences between the observed and expected counts are too large to be attributed to chance. It is on this basis (p< 0.05) that the decision to reject the hypothesis of a 1:1:1:1 ratio is made.

False df increases with increasing n (number of categories), but not necessarily with increasing number of subjects.

b) The value of observed - expected for this cell = -2. With a total of 200, the expected number in each cell when the predicted ratio is 1:1:1:1 = 50. The observed number is 48.

a) Proband The proband is the affected individual who is first brought to the attention of a medical researcher; usually the pedigree is constructed around this individual.

a) If neither parent expresses the trait, but the offspring does, both parents must be heterozygous for the trait. For an autosomal recessive trait to be expressed, the individual must be homozygous for the recessive allele. The only way (excluding new mutations) for homozygous recessive offspring to be produced from parents who do not express the trait is if both parents are heterozygous for the trait.

d) Rare X-linked recessive This trait is X-linked, and since only half the sons are affected, it is a recessive trait and the mother must have been heterozygous.

True One map unit is equal to 1% recombination between two genes; 10 map units would be equal to 10% recombination between the genes.

a) 1.25% The probability of a double crossover is the product of the probabilities of the single crossovers: 0.25 x 0.05 = 0.0125, or 1.25%.

c) 30% Recombination frequencies between linked genes along a chromosome are additive, so the recombination frequency between genes X and Z is 25 + 5 = 30.

False To construct a mapping cross of linked genes, it is important that the genotypes of all of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny, taking into consideration that the homozygote produced only recessive gametes. Gametes and their genotypes can never be observed directly.

a) 20 map units The map distance between any two genes is the sum of the percentages of all detectable recombination events between them, so 17 + 3 = 20.

b) 0.2 The coefficient of coincidence is 0.01/0.0125 = 0.8, so the interference is 1 - 0.8 = 0.2.

True Frac NCO refers to the fraction of gametes that have not undergone crossing over and thus their genotypes reflect those of the parental gametes.

c) Interference will be a significant factor in the number of crossovers observed. Interference effects are more likely when crossovers are confined to a small region.

a) The distance between A and B is less than the distance between B and C. The frequency of crossover events decreases as the distance between genes decreases.

d) some of their male progeny may have white eyes If the female is heterozygous, approximately half of the male progeny will have white eyes.

True Males inherit only one X chromosome. That chromosome is contributed by the female parent.

c) His mother had at least one white allele. Because this male had white eyes, he must have inherited a white allele from his mother.

b) X-linked dominant If the skin condition is caused by an X-linked dominant allele, a father would pass the allele on to all of his daughters, who would all have the skin condition. In contrast, the father would not pass the allele on to any of his sons because the sons would receive the father's Y chromosome, not his X chromosome. As a result, none of the sons would inherit the skin condition.

False The cell cycle does consist of two phases, but mitosis is the shorter phase. Interphase is composed of the G1, S, and G2 stages, which together are much longer than mitosis.

b) Centriole Centrioles are thought to organize the spindle fibers in animal cells, but they are not found in many plant, fungal, or algal cells.

a) Sister chromatids are aligned along the equatorial plane of the cell. During metaphase, sister chromatids are aligned along the equatorial plane, or metaphase plate, of the cell.

c) . . . G1 - S - G2 - M - G1 . . . The S phase is both preceded and followed by a period of growth (G1 and G2, respectively). After the M phase, the cell re-enters the G1 phase.

False Both haploid and diploid cells can undergo mitosis.

a) polar microtubules contract, pulling attached chromosomes toward the poles This describes the role of kinetochore microtubules during anaphase. Polar microtubules lengthen during anaphase, causing the cell to become elliptical.

False A dyad is composed of two sister chromatids joined at a common centromere.

a) 8 Meiosis reduces the number of chromosomes passed on to daughter cells by one-half.

d) The second meiotic division is similar to mitosis in that the sister chromatids separate. The second meiotic division is also called an equational division and involves the separation of sister chromatids, similar to what occurs in mitosis.

b) Pachynema Crossing over occurs during pachynema when bivalents are closely paired.

True Chromosomes are duplicated during interphase; at synapsis of prophase I, one chromosome (with two chromatids) in a tetrad is paternally inherited while the other is maternally inherited.

b) Anaphase II Sister chromatids from each dyad separate during anaphase II.

d) P site The initiator tRNA^fmet binds to the mRNA codon in the P site of the ribosome. The initiator tRNA is the only one that binds in the P site; all other tRNAs bind the ribosome in the A site.

False Peptidyl transferase is the enzyme that catalyzes the formation of peptide bonds during translation. EF‑Tu is an elongation factor that facilitates the entry of charged tRNAs into the A site.

c) mRNA shifts in the 5' direction along the ribosome. Translocation is the process by which mRNA shifts by 3 bases in the 5' direction along the ribosome to bring another codon into the A site.

e) the specific sequence of bases along the DNA strands In eukaryotes, binding of RNA polymerase II to DNA involves several other proteins known as transcription factors. Many of these transcription factors bind to the DNA in the promoter region (shown below in green), located at the 3' end of the sequence on the template strand. Although some transcription factors bind to both strands of the DNA, others bind specifically to only one of the strands. Transcription factors do not bind randomly to the DNA. Information about where each transcription factor binds originates in the base sequence to which each transcription factor binds. The positioning of the transcription factors in the promoter region determines how the RNA polymerase II binds to the DNA and in which direction transcription will occur.

c) Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes. d) A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA. e) A cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA. Once RNA polymerase II is bound to the promoter region of a gene, transcription of the template strand begins. As transcription proceeds, three key steps occur on the RNA transcript: Early in transcription, when the growing transcript is about 20 to 40 nucleotides long, a modified guanine nucleotide is added to the 5' end of the transcript, creating a 5' cap. Introns are spliced out of the RNA transcript by spliceosomes, and the exons are joined together, producing a continuous coding region. A poly-A tail (between 50 and 250 adenine nucleotides) is added to the 3' end of the RNA transcript. Only after all these steps have taken place is the mRNA complete and capable of exiting the nucleus. Once in the cytoplasm, the mRNA can participate in translation.

a) heat-killed bacteria harbor the constituent(s) necessary to convey genetic properties to living bacteria Because some of the nonvirulent bacteria acquired properties of the virulent bacteria, instructions for this transformation must be carried by the virulent bacteria.

d) incubated nonvirulent cells with the complete extract The complete extract possessed the same ability to induce transformation in IIR bacteria as whole heat-killed IIS bacteria.

b) polysaccharides are the genetic material Failure to transform suggests that the chemical degraded in that preparation is the one responsible for transformation, in this case polysaccharides.

b) Chromatids of homologous chromosomes Chromatids of homologous chromosomes can recombine during meiosis.

True The crossbridge DNA structure formed after the initial nick is sealed can migrate along the chromatid. This process is called branch migration, and it increases the length of heteroduplex DNA.

b) DNA polymerization Recombination does not include the synthesis of new DNA.

e) deoxyribose, phosphate group, thymine DNA and RNA have similar structures: a pentose sugar with a nitrogenous base and a phosphate group. DNA and RNA differ in the type of pentose sugar each possesses (DNA has deoxyribose; RNA has ribose) and in one base (DNA has thymine; RNA has uracil).

True Guanine and adenine are indeed purines found in DNA; thymine and cytosine are the pyrimidines found in DNA.

c) The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions. This statement is true; the 5′-3′ orientation of each chain runs in opposite directions.

b) 5′ CAGTCAAGCAT 3′ This sequence is complementary and in the correct orientation.

a) It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein.

a) DNA is the identity of the hereditary material in phage T2. Because phage DNA and not protein was associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material.

a) Both preparations of infected bacteria would exhibit radioactivity. Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.

False The data obtained from the Meselson-Stahl experiment after one generation was consistent with both the semiconservative and the dispersive model of DNA replication. The conservative model of DNA replication was eliminated because it predicted that there would be two bands representing the original DNA at one density and the newly replicated DNA at a different density.

d) DNA polymerase DNA polymerase catalyzes the addition of nucleotides to a growing DNA chain.

d) Single‑strand binding proteins stabilize the open conformation of the unwound DNA. Once helicase unwinds the double helix, single‑strand binding proteins bind to the open DNA and prevent it from winding together again.

False The degeneracy of the genetic code means that an amino acid may be coded for by more than one codon. However, a single codon can only ever specify one amino acid.

a) An addition and a deletion mutation A single addition or deletion would change the reading frame of the protein, but if another mutation occurred to cancel the effects of the first mutation, only those amino acids between the mutations would change.

b) A polycistronic mRNA may be transcribed if the gene products are used in the same pathway or needed at the same time. This statement is false. Polycistronic mRNAs are produced only in prokaryotes. In eukaryotes, a single gene is transcribed at a time.

False The Ames test is used as a preliminary screening tool. Not all compounds that give a positive Ames test are carcinogenic.

c) Liver enzymes may activate some innocuous compounds, making them mutagenic. Some compounds are innocuous until they are activated metabolically by liver enzymes.

a) his+ prototrophs The bacteria used in the Ames test to evaluate mutagenicity are his− auxotrophs. If the Ames test is positive, these bacteria have reverted back to wild type and are his+ prototrophs

a) Regulatory proteins that bind to specific DNA sequences Transcription factors are proteins that bind to specific DNA recognition sequences in the promoter or enhancer elements of a gene.

b) Gal4p Gal4p is a yeast protein that contains a DNA binding domain.

True Alternative splicing is a form of mRNA processing and occurs after mRNA synthesis.

False There are many VNTR alleles for each locus but not an unlimited number. People who are closely related are more likely to have similar DNA fingerprints.

c) They increase the probability of producing a DNA fingerprint that is unique to an individual. The use of a large number of probes increases the chances that the DNA fingerprint produced will be unique.

c) 50 The frequency of occurrence for the four alleles is 1 in (10 × 20 × 50 × 500) = 1 in 5,000,000, and the number of people in the population who have this pattern is 250,000,000/5,000,000 = 50.

a) They provide a defense mechanism against infection by viruses. Restriction endonucleases recognize and degrade viral DNA, thus preventing viral infections.

False Restriction endonucleases cut DNA at specific sequences, but DNA ligase must be used to bond two strands covalently with the same "sticky ends."

d) 3′ TCTTAAG 5′ This sequence does not contain the BamHI recognition site.

True Phage vectors can carry DNA fragments of about 20 kb, whereas plasmids can only carry DNA of less than 15 kb.

d) Lambda arms Lambda arms are regions that flank the inserted foreign DNA in phage λ vectors.

a) White colonies that are resistant to tetracycline The presence of blue colonies means that the plasmid taken up by these bacteria is recombinant, since the lacZ gene was disrupted.

d) Ligase Ligase is not required for a PCR reaction. The enzyme used during PCR is a thermostable DNA polymerase.

False The annealing phase takes place at the lowest temperature of PCR. Taq polymerase is derived from bacteria that live in hot springs, so the enzyme is thermostable, meaning that its enzymatic properties can withstand the high temperatures needed for denaturation.

d) To separate the double‑stranded DNA The temperature is raised to denature the double‑stranded DNA molecule into single strands.

True Restriction digestion produces fragments of DNA, and the sizes of these fragments can be determined by gel electrophoresis using standard DNA fragments of known size.

b) One 1.5‑kb fragment and one 5.5‑kb fragment EcoRI digestion will produce two fragments corresponding in size to the 1.5‑kb fragment cloned into the plasmid plus the 5.5‑kb plasmid itself.

c) The XhoI site is located within the BamHI 400‑bp fragment. Since double digestion produces a 100‑bp fragment and a 300‑bp fragment, the XhoI site must be located within the BamHI 400‑bp fragment.

c) They have a hydrogen at the 3′ carbon of the sugar. ddNTPs terminate synthesis because there is no 3′‑hydroxyl group onto which DNA polymerase can add nucleotides.

False Small DNA fragments have less hindrance in moving through the gel, so they migrate more quickly than larger fragments.

a) The DNA sequence obtained is complementary to the template strand. The DNA fragments produced in sequencing reactions are synthesized by DNA polymerase to be complementary to the template strand.

c) No genetic or physical maps of the genome are needed to begin shotgun cloning. Shotgun cloning randomly sequences clones with no prior knowledge of their location in the genome. 21.1 Whole-Genome Shotgun Sequencing Is a Widely Used Method for Sequencing and Assembling Entire Genomes (p.532)

c) identify open reading frames. Cytosine and guanine doublets are concentrated near gene-rich regions of the genome. Codon bias is the phenomenon often observed in coding regions of DNA for certain codons to be used over others coding for the same amino acid. 21.2 DNA Sequence Analysis Relies on Bioinformatics and Genomic Databases (p.536)

True Because orthologs are descended from a common ancestral gene, they tend to have the same function in different species. If the function of a gene in one species were known, then its ortholog in a different species would most likely have that same function. 21.3 Functional Genomics Attempts to Identify Potential Functions of Genes and Other Elements in a Genome (p. 540)

c) Human genes are similar in size to those of invertebrates with similar sized introns. Human genes are larger than invertebrates with larger introns. The Human Genome Project Reveals Many Important Aspects of Genome Organization in Humans (p.541)

c) gene duplication and divergence. The ancestral human globin gene was probably duplicated into two sister genes about 800 million years ago, which then diverged to give rise to the present-day genes. 21.7 Comparative Genomics Analyzes and Compares Genomes from Different Organisms (p. 550)

False The sum of the allele frequencies must equal 1, so the frequency of the N allele must be 0.35.

a) 0.043 If q^2 = 0.0005, then q = 0.022 and p = 1 − q = 0.978. The heterozygote frequency is 2pq, or 2 (0.978) (0.022) = 0.043.

a) Disruptive Birds with an intermediate beak phenotype are at a disadvantage in this population and will be selected against because they are less fit.

d) If selection can be quantified, relative fitness values can be calculated and used to appropriately modify the Hardy-Weinberg equation. Although Hardy-Weinberg predictions assume no selection, selection values, if known, can be used to modify predictions of the model.

d) The frequency of the recessive allele is 0.7.

c) The stronger the selection against a given genotype, the lower the fitness associated with that genotype. Fitness = 1 - selection.