front 1 When Thomas Hunt Morgan crossed his red-eyed F₁ generation flies to
each other, the F₂ generation included both red- and white-eyed flies.
Remarkably, all the white-eyed flies were male. What was the
explanation for this result? | back 1 B) The gene involved is on the X chromosome. |
front 2 Which of the following is the meaning of the chromosome theory of
inheritance as expressed in the early 20th century? B) Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis. C) No more than a single pair of chromosomes can be found in a healthy normal cell. D) Natural selection acts on certain chromosome arrays rather than on genes. | back 2 B) Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis. |
front 3 Males are more often affected by sex-linked traits than females
because B) female hormones such as estrogen often compensate for the effects of mutations on the X chromosome. C) X chromosomes in males generally have more mutations than X chromosomes in females. D) males are hemizygous for the X chromosome. E) mutations on the Y chromosome often worsen the effects of X-linked mutations. | back 3 D) males are hemizygous for the X chromosome. |
front 4 SRY is best described in which of the following ways? B) an autosomal gene that is required for the expression of genes on the Y chromosome C) a gene region present on the Y chromosome that triggers male development D) an autosomal gene that is required for the expression of genes on the X chromosome E) a gene required for development, and males or females lacking the gene do not survive past early childhood | back 4 C) a gene region present on the Y chromosome that triggers male development |
front 5 When Thomas Hunt Morgan crossed his red-eyed F₁ generation flies to
each other, the F₂ generation included both red- and white-eyed flies.
Remarkably, all the white-eyed flies were male. What was the
explanation for this result? | back 5 B) The gene involved is on the X chromosome. |
front 6 Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents? A) XnXn and XnY B) XnXn and XNY C) XNXN and XnY D) XNXN and XNY E) XNXn and XNY | back 6 E) XNXn and XNY |
front 7 Cinnabar eyes is a sex-linked recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a wild-type male, what percentage of the F₁ males will have cinnabar eyes? A) 0% B) 25% C) 50% D) 75% E) 100% | back 7 E) 100% |
front 8 Normally, only female cats have the tortoiseshell phenotype because_______. A) a male inherits only one allele of the X-linked gene controlling hair color. B) only males can have Barr bodies C) the Y chromosome has a gene blocking orange coloration D) multiple crossovers on the Y chromosome prevent orange pigment production. | back 8 A) a male inherits only one allele of the X-linked gene controlling hair color. |
front 9 In birds, sex is determined by a ZW chromosome scheme. Males are ZZ and females are ZW. A recessive lethal allele that causes death of the embryo is sometimes present on the Z chromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male that is heterozygous for the lethal allele and a normal female? A) 2:1 male to female B) 1:2 male to female C) 1:1 male to female D) 4:3 male to female E) 3:1 male to female | back 9 A) 2:1 male to female |
front 10 A man who carries an allele of an X-linked gene will pass it on to ___. A) Half of his daughters B) All of his sons C) All of his daughters D) All of his children | back 10 C) All of his daughters |
front 11 A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was six feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. How many of their daughters might be expected to be color-blind dwarfs? A) three out of four B) none C) one out of four D) half | back 11 B) none |
front 12 A recessive allele on the X chromosome is responsible for red-green color blindness in humans. A woman with normal vision whose father is color blind marries a color-blind male. What is the probability that this couple's first son will be color blind? A) 1/4 B) 3/4 C) 1/2 D) 2/3 | back 12 C) 1/2 |
front 13 In a Drosophila experiment, a cross was made between homozygous wild-type females and yellow-bodied males. All of the resulting F1s were phenotypically wild type. However, adult flies of the F2 generation (resulting from matings of the F1s) had the characteristics shown in the figure above. Consider the following questions: (a) Is the mutant allele for yellow body recessive or dominant? (b) Is the yellow locus autosomal (not X-linked) or X-linked? A) (a) recessive; (b) X-linked B) (a) recessive; (b) not X-linked C) (a) dominant; (b) X-linked D) (a) dominant; (b) not X-linked | back 13 A) (a) recessive; (b) X-linked |
front 14 What is the definition of one map unit? A) the recombination frequency between two genes assorting independently. B) 1 nanometer of distance between two genes C) a 1% frequency of recombination between two genes. D) the physical distance between two linked genes. | back 14 C) a 1% frequency of recombination between two genes. |
front 15 The greatest distance among the three genes is between a and c. What does this mean? A) Gene c is between a and B) Genes are in the order: a—b—c. C) Gene a is not recombining with D) Gene a is between b and | back 15 B) Genes are in the order: a—b—c. |
front 16 If cell X enters meiosis, and nondisjunction of one chromosome occurs in one of its daughter cells during meiosis II, what will be the result at the completion of meiosis? A) All the gametes descended from cell X will be diploid. B) Half of the gametes descended from cell X will be n + 1, and half will be n - 1. C) 1/4 of the gametes descended from cell X will be n + 1, 1/4 will be n - 1, and 1/2 will be n. D) Two of the four gametes descended from cell X will be haploid, and two will be diploid. | back 16 C) 1/4 of the gametes descended from cell X will be n + 1, 1/4 will be n - 1, and 1/2 will be n. |
front 17 Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual? A) 47, trisomy 21 B) 47, XXY C) 47, XXX D) 47, XYY E) 45, X | back 17 A) 47, trisomy |
front 18 A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition? A) The woman inherited this tendency from her parents. B) The mother had a chromosomal duplication. C) One member of the couple underwent nondisjunction in somatic cell production. D) The mother most likely underwent nondisjunction during gamete production. | back 18 D) The mother most likely underwent nondisjunction during gamete production. |
front 19 Correns described that the inheritance of variegated color on the leaves of certain plants was determined by the maternal parent only. What phenomenon does this describe? A) mitochondrial inheritance B) chloroplast inheritance C) genomic imprinting D) infectious inheritance E) sex-linkage | back 19 B) chloroplast inheritance |
front 20 Mitochondrial DNA is primarily involved in coding for proteins needed for protein complexes of the electron transport chain and ATP synthase. Therefore, mutations in mitochondrial genes would most affect_____. A) DNA synthesis in cells of the immune system B) the movement of oxygen into erythrocytes C) the storage of urine in the urinary bladder D) the generation of ATP in muscle cells | back 20 C) the storage of urine in the urinary bladder |
front 21 The pedigree in the figure above shows the transmission of a trait in a particular family. Based on this pattern of transmission, the trait is most likely _____. A) mitochondrial B) sex-linked dominant C) sex-linked recessive D) autosomal dominant | back 21 A) mitochondrial |
front 22 During meiosis, a defect occurs in a cell that results in the failure of microtubules, spindle fibers, to bind at the kinetochores, a protein structure on chromatids where the spindle fibers attach during cell division to pull sister chromatids apart. Which of the following is the most likely result of such a defect? A) New microtubules with more effective binding capabilities to kinetochores will be synthesized to compensate for the defect. B) Excessive cell divisions will occur resulting in cancerous tumors and an increase in the chromosome numbers known as polyploidy. C) The defect will be bypassed in order to and ensure normal chromosome distribution in the new cells. D) The resulting cells will not receive the correct number of chromosomes in the gametes, a condition known as aneuploidy. | back 22 D) The resulting cells will not receive the correct number of chromosomes in the gametes, a condition known as aneuploidy. |
front 23 Inheritance patterns cannot always be explained by Mendel’s models of inheritance. If a pair of homologous chromosomes fails to separate during meiosis I, select the choice that shows the chromosome number of the four resulting gametes with respect to the normal haploid number (n)? A) n+1; n+1; n-1; n-1 B) n+1; n-1; n; n C) n+1; n-1; n-1; n-1 D) n+1; n+1; n; n | back 23 A) n+1; n+1; n-1; n-1 |
front 24 In his transformation experiments, what did Griffith observe? A) Mutant mice were resistant to bacterial infections. B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form. C) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic. D) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains. E) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice. | back 24 B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form. |
front 25 Which of the following investigators was/were responsible for the following discovery? In DNA from any species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine. A) Frederick Griffith B) Alfred Hershey and Martha Chase C) Oswald Avery, Maclyn McCarty, and Colin MacLeod D) Erwin Chargaff E) Matthew Meselson and Franklin Stahl | back 25 D) Erwin Chargaff |
front 26 For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work? A) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins B) Although there are more nitrogens in a nucleotide, labeled phosphates actually have sixteen extra neutrons; therefore, they are more radioactive. C) There is no radioactive isotope of nitrogen D) Radioactive nitrogen has a half-life of 100,000 years, and he material would be too dangerous for too long | back 26 A) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins |
front 27 In the polymerization of DNA, a phosphodiester bond is formed between a phosphate group of the nucleotide being added and _____ of the last nucleotide in the polymer. A) the 5' phosphate B) C6 C) the 3' OH D) a nitrogen from the nitrogen-containing base | back 27 C) the 3' OH |
front 28 At a specific area of a chromosome, the sequence of nucleotides below
is present where the chain opens to form a replication fork: A) 5' G C C T A G G 3' B) 3' G C C T A G G 5' C) 5' A C G T T A G G 3' D) 5' A C G U U A G G 3' E) 5' G C C U A G G 3' | back 28 D) 5' A C G U U A G G 3' |
front 29 The difference between ATP and the nucleoside triphosphates used during DNA synthesis is that A) the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose. B) the nucleoside triphosphates have two phosphate groups; ATP has three phosphate groups. C) ATP contains three high-energy bonds; the nucleoside triphosphate have two. D) ATP is found only in human cells; the nucleoside triphosphates are found in all animal and plant cells. E) triphosphate monomers are active in the nucleoside triphosphates, but not in ATP. | back 29 A) the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose. |
front 30 Which of the following help(s) to hold the DNA strands apart while they are being replicated? A) primase B) ligase C) DNA polymerase D) single-strand binding proteins E) exonuclease | back 30 D) single-strand binding proteins |
front 31 In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (¹⁵N) and then transferred them to a medium containing ¹⁴N. Which of the results in the figure above would be expected after one round of DNA replication in the presence of ¹⁴N? A) A B) B C) C D) D E) E | back 31 D) D |
front 32 A space probe returns with a culture of a microorganism found on a distant planet. Analysis shows that it is a carbon-based life-form that has DNA. You grow the cells in ¹⁵N medium for several generations and then transfer them to ¹⁴N medium. Which pattern in the figure above would you expect if the DNA was replicated in a conservative manner? A) A B) B C) C D) D E) E | back 32 B) B |
front 33 After the first replication was observed in their experiments testing the nature of DNA replication, Meselson and Stahl could be confident of which of the following conclusions? A) Replication is semi-conservative. B) Replication is not dispersive. C) Replication is not semi-conservative. D) Replication is not conservative. E) Replication is neither dispersive nor conservative. | back 33 D) Replication is not conservative. |
front 34 Semiconservative replication involves a template. What is the template? A) one strand of the DNA molecule B) an RNA molecule C) single-stranded binding proteins D) DNA polymerase | back 34 A) one strand of the DNA molecule |
front 35 What is the difference between the leading strand and the lagging strand in DNA replication? A) There are different DNA polymerases involved in the elongation of the leading strand and the lagging strand. B) The leading strand is synthesized continuously in the 5' -> 3' direction, while the lagging strand is synthesized discontinuously in the 5' -> 3' direction. C) The leading strand requires an RNA primer, whereas the lagging strand does not. D) The leading strand is synthesized in the 3' -> 5' direction in a discontinuous fashion, while the lagging strand is synthesized in the 5' -> 3' direction in a continuous fashion. | back 35 B) The leading strand is synthesized continuously in the 5' -> 3' direction, while the lagging strand is synthesized discontinuously in the 5' -> 3' direction. |
front 36 What is a major difference between eukaryotic DNA replication and prokaryotic DNA replication? A) DNA polymerases of prokaryotes can add nucleotides to both 3' and 5' ends of DNA strands, while those of eukaryotes function only in the 5' -> 3' direction B) Prokaryotic replication does not require a primer C) DNA replication in prokaryotic cells is conservative. DNA replication is eukaryotic cells is semi-conservative. D) Prokaryotic chromosomes have a single origin of replication, while eukaryotic chromosomes have multiple origins of replication. | back 36 D) Prokaryotic chromosomes have a single origin of replication, while eukaryotic chromosomes have multiple origins of replication. |
front 37 Telomere shortening puts a limit on the number of times a cell can divide. Research has shown that telomerase can extend the life span of cultured human cells. How might adding telomerase affect cellular aging? A) Telomerase will speed up the rate of cell proliferation. B) Telomerase eliminates telomere shortening and retards aging. C) Telomerase shortens telomeres, which delays cellular aging. D) Telomerase would have no effect on cellular aging. | back 37 B) Telomerase eliminates telomere shortening and retards aging. |
front 38 Which of the following statements describes the eukaryotic chromosome? A) It is composed of DNA alone. B) The nucleosome is its most basic functional subunit. C) The number of genes on each chromosome is different in different cell types of an organism. D) a single linear molecule of double-stranded DNA plus proteins. E) Active transcription occurs on heterochromatin but not euchromatin. | back 38 D) a single linear molecule of double-stranded DNA plus proteins. |
front 39 If a cell were unable to produce histone proteins, which of the
following would be a likely effect? B) The cell's DNA couldn't be packed into its nucleus. C) Spindle fibers would not form during prophase. D) Amplification of other genes would compensate for the lack of histones. E) Pseudogenes would be transcribed to compensate for the decreased protein in the cell. | back 39 B) The cell's DNA couldn't be packed into its nucleus. |
front 40 Which of the following represents the order of increasingly higher levels of organization of chromatin? A) nucleosome, 30-nm chromatin fiber, looped domain B) looped domain, 30-nm chromatin fiber, nucleosome C) looped domain, nucleosome, 30-nm chromatin fiber D) nucleosome, looped domain, 30-nm chromatin fiber E) 30-nm chromatin fiber, nucleosome, looped domain | back 40 A) nucleosome, 30-nm chromatin fiber, looped domain |
front 41 In E. coli replication the enzyme primase is used to attach a 5 to 10 base ribonucleotide strand complementary to the parental DNA strand. The RNA strand serves as a starting point for the DNA polymerase that replicates the DNA. If a mutation occurred in the primase gene, which of the following would you expect? A) Replication would only occur on the leading strand. B) Replication would only occur on the lagging strand. C) Replication would not occur on either the leading or lagging strand. D) Replication would not be affected as the enzyme primase in involved with RNA synthesis. | back 41 C) Replication would not occur on either the leading or lagging strand. |
front 42 Garrod hypothesized that "inborn errors of metabolism" such as alkaptonuria occur because A) metabolic enzymes require vitamin cofactors, and affected individuals have significant nutritional deficiencies. B) enzymes are made of DNA, and affected individuals lack DNA polymerase. C) many metabolic enzymes use DNA as a cofactor, and affected individuals have mutations that prevent their enzymes from interacting efficiently with DNA. D) certain metabolic reactions are carried out by ribozymes, and affected individuals lack key splicing factors. E) genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes. | back 42 E) genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes. |
front 43 The genetic code is essentially the same for all organisms. From this, one can logically assume which of the following? A) A gene from an organism can theoretically be expressed by any other organism. B) All organisms have experienced convergent evolution. C) DNA was the first genetic material. D) The same codons in different organisms translate into the different amino acids. E) Different organisms have different numbers of different types of amino acids. | back 43 A) A gene from an organism can theoretically be expressed by any other organism. |
front 44 A possible sequence of nucleotides in the template strand of DNA that would code for the polypeptide sequence phe-leu-ile-val would be A) 5' TTG-CTA-CAG-TAG 3'. B) 3' AAC-GAC-GUC-AUA 5'. C) 5' AUG-CTG-CAG-TAT 3'. D) 3' AAA-AAT-ATA-ACA 5'. E) 3' AAA-GAA-TAA-CAA 5'. | back 44 D) 3' AAA-AAT-ATA-ACA 5'. |
front 45 What amino acid sequence will be generated, based on the following
mRNA codon sequence? A) met-arg-glu-arg-glu-arg B) met-glu-arg-arg-glu-leu C) met-ser-leu-ser-leu-ser D) met-ser-ser-leu-ser-leu E) met-leu-phe-arg-glu-glu | back 45 D) met-ser-ser-leu-ser-leu |
front 46 According to the central dogma, what molecule should go in the blank? DNA -> ____ -> Proteins A) tRNA B) mtDNA C) rRNA D) mRNA | back 46 D) mRNA |
front 47 Which of the following statements best describes the termination of transcription in prokaryotes? A) RNA polymerase transcribes through the polyadenylation signal, causing proteins to associate with the transcript and cut it free from the polymerase. B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to separate from the DNA and release the transcript. C) RNA polymerase transcribes through an intron, and the snRNPs cause the polymerase to let go of the transcript. D) Once transcription has initiated, RNA polymerase transcribes until it reaches the end of the chromosome. E) RNA polymerase transcribes through a stop codon, causing the polymerase to stop advancing through the gene and release the mRNA. | back 47 B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to separate from the DNA and release the transcript. |
front 48 Alternative RNA splicing ________. A) can allow the production of similar proteins from different RNAs B) can allow the production of proteins of different sizes and functions from a single mRNA C) is a mechanism for increasing the rate of translation D) increases the rate of transcription | back 48 B) can allow the production of proteins of different sizes and functions from a single mRNA |
front 49 Which of the following statements best describes the termination of transcription in prokaryotes? A) RNA polymerase transcribes through the polyadenylation signal, causing proteins to associate with the transcript and cut it free from the polymerase. B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to separate from the DNA and release the transcript. C) RNA polymerase transcribes through an intron, and the snRNPs cause the polymerase to let go of the transcript. D) Once transcription has initiated, RNA polymerase transcribes until it reaches the end of the chromosome. E) RNA polymerase transcribes through a stop codon, causing the polymerase to stop advancing through the gene and release the mRNA. | back 49 B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to separate from the DNA and release the transcript. |
front 50 In an experimental situation, a student researcher inserts an mRNA molecule into a eukaryotic cell after he has removed its 5' cap and poly-A tail. Which of the following would you expect him to find? A) The mRNA could not exit the nucleus to be translated. B) The cell recognizes the absence of the tail and polyadenylates the mRNA. C) The molecule is digested by restriction enzymes in the nucleus. D) The molecule is digested by exonucleases since it is no longer protected at the 5' end. E) The molecule attaches to a ribosome and is translated, but more slowly. | back 50 D) The molecule is digested by exonucleases since it is no longer protected at the 5' end. |
front 51 5' UTR E₁ I₁ E₂ I₂ E₃ I₃ E₄ UTR 3' Which components of the previous molecule will also be found in mRNA in the cytosol? A) 5' UTR I₁ I₂ I₃ UTR 3' B) 5' E₁ E₂ E₃ E₄ 3' C) 5' UTR E₁ E₂ E₃ E₄ UTR 3' D) 5' I₁ I₂ I₃ 3' E) 5' E₁ I₁ E₂ I₂ E₃ I₃ E₄ 3' | back 51 C) 5' UTR E₁ E₂ E₃ E₄ UTR 3' |
front 52 A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is A) TTT B) UUA C) UUU D) AAA E) either UAA or TAA, depending on first base wobble | back 52 D) AAA |
front 53 Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the A) binding of ribosomes to mRNA. B) shape of the A and P sites of ribosomes. C) bonding of the anticodon to the codon. D) attachment of amino acids to tRNAs. E) bonding of the anticodon to the codon and the attachment of amino acids to tRNAs. | back 53 E) bonding of the anticodon to the codon and the attachment of amino acids to tRNAs. |
front 54 Which one of the following statements about RNA processing is true? A) Exons are cut out before mRNA leaves the nucleus. B) Ribozymes may function in RNA splicing. C) RNA splicing can be catalyzed by tRNA. D) A primary transcript is often much shorter than the final RNA molecule that leaves the nucleus. | back 54 B) Ribozymes may function in RNA splicing. |
front 55 A primary transcript in the nucleus of a eukaryotic cell is _____ the functional mRNA, while a primary transcript in a prokaryotic cell is _____ the functional mRNA. A) the same size as; smaller than B) larger than; the same size as C) larger than; smaller than D) the same size as; larger than | back 55 B) larger than; the same size as |
front 56 In the structural organization of many eukaryotic genes, individual exons may be related to which of the following? A) the sequence of the intron that immediately precedes each exon B) the number of polypeptides making up the functional protein C) the various domains of the polypeptide product D) the number of start sites for transcription | back 56 C) the various domains of the polypeptide product |
front 57 In eukaryotes there are several different types of RNA polymerase. Which type is involved in transcription of mRNA for a globin protein? A) ligase B) RNA polymerase I C) RNA polymerase II D) RNA polymerase III E) primase | back 57 C) RNA polymerase II |
front 58 Transcription in eukaryotes requires which of the following in addition to RNA polymerase? A) the protein product of the promoter B) start and stop codons C) ribosomes and tRNA D) several transcription factors (TFs) E) aminoacyl synthetase | back 58 D) several transcription factors (TFs) |
front 59 Which of the following best describes the significance of the TATA box in eukaryotic promoters? A) It sets the reading from of the mRNA B) It is the recognition site for ribosomal binding C) Its significance has not yet been determined D) It is the recognition site for a specific transcription factor | back 59 D) It is the recognition site for a specific transcription factor |
front 60 Which of the following does not occur in prokaryotic gene expression, but does occur in eukaryotic gene expression? A) mRNA, tRNA, and rRNA are transcribed B) RNA polymerase binds to the promoter C) RNA polymerase requires a primer to elongate the molecule D) A cap is added to the 5' end of the mRNA | back 60 D) A cap is added to the 5' end of the mRNA |
front 61 A ribozyme is _____. A) a catalyst that uses RNA as a substrate B) an enzyme that catalyzes the association between the large and small ribosomal subunits C) an RNA with catalytic activity D) an enzyme that synthesizes RNA as a part of the transcription process | back 61 C) an RNA with catalytic activity |
front 62 Codons are three-base sequences that specify the addition of a single amino acid. How do eukaryotic codons and prokaryotic codons compare? A) Prokaryotic codons usually contain different bases than those of eukaryotes. B) Prokaryotic codons usually specify different amino acids than those of eukaryotes. C) The translation of codons is mediated by tRNAs in eukaryotes, but translation requires no intermediate molecules such as tRNAs in prokaryotes. D) Codons are a nearly universal language among all organisms. | back 62 D) Codons are a nearly universal language among all organisms. |
front 63 Which of the following occurs in prokaryotes but not in eukaryotes? A) translation in the absence of a ribosome B) post-transcriptional splicing C) concurrent transcription and translation D) gene regulation | back 63 C) concurrent transcription and translation |
front 64 Refer to the figure above. What would the anticodon be for a tRNA that transports phenylalanine to a ribosome? A) UUU B) AAA C) TTT D) CCC | back 64 B) AAA |
front 65 Which of the following contradicts the one-gene, one-enzyme hypothesis? A) A mutation in a single gene can result in a defective protein. B) Alkaptonuria results when individuals lack a single enzyme involved in the catalysis of homogentisic acid. C) Sickle-cell anemia results in defective hemoglobin. D) A single antibody gene can code for different related proteins, depending on the splicing that takes place post-transcriptionally. | back 65 D) A single antibody gene can code for different related proteins, depending on the splicing that takes place post-transcriptionally. |
front 66 Which of the following is directly related to a single amino acid? A) the base sequence of the tRNA B) the amino acetyl tRNA synthase C) the three-base sequence of mRNA D) the complementarity of DNA and RNA | back 66 C) the three-base sequence of mRNA |
front 67 In the process of transcription, _____. A) DNA is replicated B) RNA is synthesized C) proteins are synthesized D) mRNA attaches to ribosomes | back 67 B) RNA is synthesized |
front 68 Codons are part of the molecular structure of _____. A) a protein B) mRNA C) tRNA D) rRNA | back 68 B) mRNA |
front 69 What does it mean when we say the genetic code is redundant? A) A single codon can specify the addition of more than one amino acid. B) The genetic code is different for different domains of organisms. C) The genetic code is universal (the same for all organisms). D) More than one codon can specify the addition of the same amino acid. | back 69 D) More than one codon can specify the addition of the same amino acid. |
front 70 Once researchers identified DNA as the unit of inheritance, they asked how information was transferred from the DNA in the nucleus to the site of protein synthesis in the cytoplasm. What is the mechanism of information transfer in eukarotes? A) DNA from a single gene is replicated and transferred to the cytoplasm, where it serves as a template for protein synthesis. B) Messenger RNA is transcribed from a single gene and transfers information from the DNA in the nucleus to the cytoplasm, where protein synthesis takes place. C) Proteins transfer information from the nucleus to the ribosome, where protein synthesis takes place. D) Transfer RNA takes information from DNA directly to a ribosome, where protein synthesis takes place. | back 70 B) Messenger RNA is transcribed from a single gene and transfers information from the DNA in the nucleus to the cytoplasm, where protein synthesis takes place. |
front 71 The figure above shows a simple metabolic pathway. According to Beadle and Tatum's hypothesis, how many genes are necessary for this pathway? A) 1 B) 2 C) 3 D) It cannot be determined from the pathway. | back 71 B) 2 |
front 72 Refer to the metabolic pathway illustrated above. If A, B, and C are all required for growth, a strain that is mutant for the gene-encoding enzyme A would be able to grow on medium supplemented with _____. A) nutrient A only B) nutrient B only C) nutrient C only D) nutrients A and C | back 72 B) nutrient B only |
front 73 Refer to the metabolic pathway illustrated above. If A, B, and C are all required for growth, a strain mutant for the gene-encoding enzyme B would be able to grow on medium supplemented with _____. A) nutrient A only B) nutrient B only C) nutrient C only D) nutrients A and C | back 73 C) nutrient C only |
front 74 A particular triplet of bases in the template strand of DNA is 5' AGT 3'. The corresponding codon for the mRNA transcribed is A) 3' UCA 5'. B) 3' UGA 5'. C) 5' TCA 3'. D) 3' ACU 5'. | back 74 A) 3' UCA 5'. |
front 75 Which of the following statements describes chromatin? A) Heterochromatin is composed of DNA, whereas euchromatin is made of DNA and RNA. B) Both heterochromatin and euchromatin are found in the cytoplasm. C) Heterochromatin is highly condensed, whereas euchromatin is less compact. D) Euchromatin is not transcribed, whereas heterochromatin is transcribed. E) Only euchromatin is visible under the light microscope. | back 75 C) Heterochromatin is highly condensed, whereas euchromatin is less compact. |
front 76 Telomere shortening is a problem in which types of cells? A) only prokaryotic cells B) only eukaryotic cells C) cells in prokaryotes and eukaryotes | back 76 B) only eukaryotic cells |
front 77 Where of the following cells have reduced or very little active telomerase activity? A) most normal germ cells B) most cancer cells C) most normal somatic cells | back 77 C) most normal somatic cells |
front 78 What is a telomere? A) the mechanism that holds two sister chromatids together B) DNA replication during telophase C) the site of origin of DNA replication D) the ends of linear chromosomes | back 78 D) the ends of linear chromosomes |
front 79 DNA is synthesized through a process known as _____. A) semiconservative replication B) conservative replication C) translation D) transcription | back 79 A) semiconservative replication |
front 80 DNA contains the template needed to copy itself, but it has no
catalytic activity in cells. What enzyme catalyzes A) ribozymes B) DNA polymerase C) deoxyribonucleotide triphosphates D) ATP | back 80 B) DNA polymerase |
front 81 What provides the energy for the polymerization reactions in DNA synthesis? A) DNA polymerase B) the deoxyribonucleotide triphosphate substrates C) breaking down the hydrogen bonds between complementary DNA strands D) ATP | back 81 B) the deoxyribonucleotide triphosphate substrates |
front 82 Refer to the figure above. What bases will be added to the primer as DNA replication proceeds? The bases should appear in the new strand in the order that they will be added starting at the 3' end of the primer. A) C, A, G, C, A, G, A B) T, C, T, G, C, T, G C) A, G, A, C, G, A, C D) G, T, C, G, T, C, T | back 82 C) A, G, A, C, G, A, C |
front 83 You briefly expose bacteria undergoing DNA replication to radioactively labeled nucleotides. When you centrifuge the DNA isolated from the bacteria, the DNA separates into two classes. One class of labeled DNA includes very large molecules (thousands or even millions of nucleotides long), and the other includes short stretches of DNA (several hundred to a few thousand nucleotides in length). These two classes of DNA probably represent A) leading strands and Okazaki fragments. B) lagging strands and Okazaki fragments. C) Okazaki fragments and RNA primers. D) leading strands and RNA primers. E) RNA primers and mitochondrial DNA. | back 83 A) leading strands and Okazaki fragments. |
front 84 Within a double-stranded DNA molecule, adenine forms hydrogen bonds with thymine and cytosine forms hydrogen bonds with guanine. This arrangement _______. A) determines the type of protein produced B) determines the tertiary structure of a DNA molecule C) permits complementary base pairing D) allows variable width of the double helix | back 84 C) permits complementary base pairing |
front 85 The leading and the lagging strands differ in that _____. A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction B) the leading strand is synthesized by adding nucleotides to the end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the end C) the lagging strand is synthesized continuously, whereas the leading strand is synthesized in short fragments that are ultimately stitched together D) the leading strand is synthesized at twice the rate of the lagging strand | back 85 A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction |
front 86 A new DNA strand elongates only in the 5' to 3' direction because _____. A) DNA polymerase begins adding nucleotides at the 5' end of the template B) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end C) replication must progress toward the replication fork D) DNA polymerase can add nucleotides only to the free 3' end | back 86 D) DNA polymerase can add nucleotides only to the free 3' end |
front 87 What is the function of topoisomerase? A) relieving strain in the DNA ahead of the replication fork B) elongating new DNA at a replication fork by adding nucleotides to the existing chain C) unwinding of the double helix D) stabilizing single-stranded DNA at the replication fork | back 87 A) relieving strain in the DNA ahead of the replication fork |
front 88 What is the role of DNA ligase in the elongation of the lagging strand during DNA replication? A) It synthesizes RNA nucleotides to make a primer. B) It joins Okazaki fragments together. C) It unwinds the parental double helix. D) It stabilizes the unwound parental DNA. | back 88 B) It joins Okazaki fragments together. |
front 89 In E. coli, to repair a thymine dimer by nucleotide excision repair, in which order do the necessary enzymes act? A) nuclease, DNA polymerase III, RNA primase B) helicase, DNA polymerase I, DNA ligase C) DNA ligase, nuclease, helicase D) nuclease, DNA polymerase I, DNA ligase | back 89 D) nuclease, DNA polymerase I, DNA ligase |
front 90 Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons? A) Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not. B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many. C) The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes. D) Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not. | back 90 B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many. |
front 91 What is meant by the description "antiparallel" regarding the strands that make up DNA? A) The twisting nature of DNA creates nonparallel strands. B) The 5' to 3' direction of one strand runs counter to the to direction of the other strand. C) Base pairings create unequal spacing between the two DNA strands. D) One strand contains only purines and the other contains only pyrimidines. | back 91 B) The 5' to 3' direction of one strand runs counter to the to direction of the other strand. |
front 92 Suppose you are provided with an actively dividing culture of E.coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base? A) Neither of the two daughter cells would be radioactive B) One of the daughter cells, but not the other, would have radioactive DNA. C) All four bases of the DNA would be radioactive. D) DNA in both daughter cells would be radioactive. | back 92 D) DNA in both daughter cells would be radioactive. |
front 93 In E.coli, there is a mutation in a gene called dnaB that alters the helicase that normally acts at the origin. Which of the following would you expect as a result of the this mutation? A) No replication fork will be formed. B) Additional proofreading will occur. C) Réplication will occur via RNA polymerase alone. D) Replication will require a DNA template from another source. | back 93 A) No replication fork will be formed. |
front 94 In E.coli, which enzyme catalyzes the elongation of a new DNA strand in the 5' --> 3' direction? A) helicase B) primase C) DNA ligase D) DNA polymerase III | back 94 D) DNA polymerase III |
front 95 Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following? A) DNA polymerase that cannot replicate the leading strand template to its 5' end B) gaps left at the 5' end of the lagging strand C) gaps left at the 3' end of the lagging strand because of the need for a primer D) the evolution of telomerase enzyme | back 95 B) gaps left at the 5' end of the lagging strand |
front 96 How does the enzyme telomerase meet the challenge of replicating the ends of linear chromosomes? A) It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity. B) It adds a single 5' cap structure that resists degradation by nucleases. C) It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity. D) It causes specific double-strand DNA breaks that result in blunt ends on both stands. | back 96 C) It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity. |
front 97 Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons? A) The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes. B) Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not. C) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many. D) Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not. | back 97 C) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many. |
front 98 Hersey and Chase set out to determine what molecule served as the unit of inheritance. They completed a series of experiments in which E.coli was infected by a T2 virus. Which molecular component of the T2 virus actually ended up inside the cell? A) ribosome B) DNA C) RNA D) protein | back 98 B) DNA |
front 99 Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine? A) 16% B) 58% C) 42% D) 8% | back 99 D) 8% |
front 100 It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following? A) sequence of bases B) phosphate-sugar backbones C) complementary pairing of bases D) side groups of nitrogenous bases E) different five-carbon sugars | back 100 A) sequence of bases |
front 101 In an analysis of the nucleotide composition of DNA, which of the following will be found? A) A = C B) A = G and C = T C) A + C = G + T D) G + C = T + A | back 101 C) A + C = G + T |
front 102 How do we describe transformation in bacteria? A) the creation of a strand of DNA from an RNA molecule B) the creation of a strand of RNA from a DNA molecule C) the infection of cells by a phage DNA molecule D) the type of semiconservative replication shown by DNA E) assimilation of external DNA into a cell | back 102 E) assimilation of external DNA into a cell |
front 103 After mixing a heat-killed, phosphorescent strain of bacteria with a living nonphosphorescent strain, you discover that some of the living cells are now phosphorescent. Which observations would provide the best evidence that the ability to fluoresce is a heritable trait? A) DNA passed from the heat-killed strain to the living strain. B) Protein passed from the heat-killed strain to the living strain. C) The phosphorescence in the living strain is especially bright. D) Phosphorescence in descendants of the living cells. | back 103 D) Phosphorescence in descendants of the living cells. |
front 104 In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts? A) DNA contains sulfur, whereas protein does not. B) DNA contains phosphorus, whereas protein does not. C) DNA contains nitrogen, whereas protein does not. D) DNA contains purines, whereas protein includes pyrimidines. | back 104 B) DNA contains phosphorus, whereas protein does not. |
front 105 What is a syndrome? A) a characteristic facial appearance B) a group of traits, all of which must be present if an aneuploidy is to be diagnosed C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation D) a characteristic trait usually given the discoverer's name E) a characteristic that only appears in conjunction with one specific aneuploidy | back 105 C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation |
front 106 One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called? A) deletion B) transversion C) inversion D) translocation E) duplication | back 106 D) translocation |
front 107 What is the reason that closely linked genes are typically inherited together? A) They are located close together on the same chromosome. B) The number of genes in a cell is greater than the number of chromosomes. C) Chromosomes are unbreakable. D) Alleles are paired together during meiosis. E) Genes align that way during metaphase I of meiosis. | back 107 A) They are located close together on the same chromosome. |
front 108 In Drosophila melanogaster, vestigial wings are caused by a recessive allele of a gene that is linked to a gene with a recessive allele that causes black body color. Morgan crossed black-bodied, normal-winged females and gray-bodied, vestigial-winged males. The F1 were all gray bodied, normal winged. The F1 females were crossed to homozygous recessive males to produce test cross progeny. Morgan calculated the map distance to be 17 map units. Which of the following is correct about the test cross progeny? A) black-bodied, normal-winged flies = 17% of the total B) gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total C) black-bodied, vestigial-winged flies = 17% of the total D) black-bodied, normal-winged flies PLUS gray-bodied, vestigial-winged flies = 17% of the total | back 108 B) gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total |
front 109 Recombination between linked genes comes about for what reason? A) Mutation on one homolog is different from that on the other homolog. B) Independent assortment sometimes fails because Mendel had not calculated appropriately. C) When genes are linked they always "travel" together at anaphase. D) Crossovers between these genes result in chromosomal exchange. E) Nonrecombinant chromosomes break and then re-join with one another. | back 109 D) Crossovers between these genes result in chromosomal exchange. |
front 110 Map units on a linkage map cannot be relied upon to calculate physical distances on a chromosome for which of the following reasons? A) The frequency of crossing over varies along the length of the chromosome. B) The relationship between recombination frequency and map units is different in every individual. C) Physical distances between genes change during the course of the cell cycle. D) The gene order on the chromosomes is slightly different in every individual. E) Linkage map distances are identical between males and females. | back 110 A) The frequency of crossing over varies along the length of the chromosome. |
front 111 | back 111 E) A and G |
front 112 Sturtevant provided genetic evidence for the existence of four pairs
of chromosomes in Drosophila in which of these ways? B) Drosophila genes cluster into four distinct groups of linked genes. C) The overall number of genes in Drosophila is a multiple of four. D) The entire Drosophila genome has approximately 400 map units. E) Drosophila genes have, on average, four different alleles. | back 112 B) Drosophila genes cluster into four distinct groups of linked genes. |
front 113 Which of the following statements is true of linkage? A) The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them. B) The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%. C) All of the traits that Mendel studied–seed color, pod shape, flower color, and others–are due to genes linked on the same chromosome. D) Linked genes are found on different chromosomes. E) Crossing over occurs during prophase II of meiosis. | back 113 A) The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them. |
front 114 How would one explain a testcross involving F₁ dihybrid flies in
which more parental-type offspring than recombinant-type offspring are
produced? B) The two genes are linked but on different chromosomes. C) Recombination did not occur in the cell during meiosis. D) The testcross was improperly performed. E) Both of the characters are controlled by more than one gene. | back 114 A) The two genes are closely linked on the same chromosome. |
front 115 A man who carries an alley of an X-linked gene will pass it on to ___. A) Half of his daughters B) All of his sons C) All of his daughters D) All of his children | back 115 C) All of his daughters |
front 116 Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as a recessive allele of an X-linked gene in humans. A woman whose father suffered from G6PD marries a normal man. i) What proportion of their sons is expected to be G6PD? ii) If the husband was not normal but was G6PD deficient, would you change your answer in part i? A) (i) 1/2 (ii) Yes B) (i) 100% (ii) No C) (i) 0 (ii) No D) (i) 1/2 (ii) No | back 116 D) (i) 1/2 (ii) No |
front 117 A mutant bacterial cell has a defective aminoacyl -tRNA synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of the normal phenylalanine. The consequence of this for the cell will be that A) none of the proteins in the cell will contain phenylalanine. B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU. C) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons. D) the ribosome will skip a codon every time a UUU is encountered. | back 117 B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU. |
front 118 There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best explained by the fact that A) some tRNAs have anticodons that recognize four or more different codons. B) the rules for base pairing between the third base of a codon and tRNA are flexible. C) many codons are never used, so the tRNAs that recognize them are dispensable. D) the DNA codes for all 61 tRNAs but some are then destroyed. E) competitive exclusion forces some tRNAs to be destroyed by nucleases. | back 118 B) the rules for base pairing between the third base of a codon and tRNA are flexible. |
front 119 The release factor (RF) ____. A) It separates tRNA in the A site from the growing polypeptide. B) It binds to the stop codon in the A site in place of a tRNA. C) It releases the amino acid from its tRNA to allow the amino acid to form a peptide bond. D) It supplies a source of energy for termination of translation. E) It releases the ribosome from the ER to allow polypeptides into the cytosol. | back 119 B) It binds to the stop codon in the A site in place of a tRNA. |
front 120 The anticodon loop of the first tRNA that will complement this mRNA is A) 3' GGC 5' B) 5' GGC 3' C) 5' ACG 3' D) 5' UGC 3' E) 3' UGC 5' | back 120 A) 3' GGC 5' |
front 121 What type of bonding is responsible for maintaining the shape of the tRNA molecule shown in the figure above? A) covalent bonding between sulfur atoms B) ionic bonding between phosphates C) hydrogen bonding between base pairs D) van der Waals interactions between hydrogen atoms E) peptide bonding between amino acids | back 121 C) hydrogen bonding between base pairs |
front 122 The figure represents tRNA that recognizes and binds a particular amino acid (in this instance, phenylalanine). Which codon on the mRNA strand codes for this amino acid? A) UGG B) GUG C) GUA D) UUC E) CAU | back 122 D) UUC |
front 123 The tRNA shown in the figure has its 3' end projecting beyond its 5' end. What will occur at this 3' end? A) The codon and anticodon complement one another. B) The amino acid binds covalently. C) The excess nucleotides (ACCA) will be cleaved off at the ribosome. D) The small and large subunits of the ribosome will attach to it. E) The 5' cap of the mRNA will become covalently bound | back 123 B) The amino acid binds covalently. |
front 124 Translation requires ____. A) mRNA, tRNA, DNA, and rRNA B) mRNA, tRNA, and rRNA C) mRNA, tRNa, and DNA D) mRNA, DNA, and rRNA | back 124 B) mRNA, tRNA, and rRNA |
front 125 Once a peptide has been formed between the amino acid attached to the tRNa in the P site and the amino acid associated with the tRNAi in the A site, what occurs next? A) initiation B) reading of the next codon of mRNA C) tranlocation D) The codon-anticodon hydrogen bonds holding the tRNA in the A site are broken | back 125 C) tranlocation |
front 126 Which one of the following, if missing, would usually prevent translation from starting? A) poly- A tail B) 5' cap C) AUG codon D) exon | back 126 C) AUG codon |
front 127 Put the following events of elongation in prokaryotic translation in chronological order. 1. Binding of mRNA with small ribosomal subunit 2. Recognition of initiation codon 3. Complementary base pairing between initiator codon and anticodon of initiator tRNA 4. Base pairing of the mRNA codon following the initiator codon with its complementary tRNA 5. Attachment of the large subunit A) 5,4,3,2,1 B) 1,2,3,5,4 C) 1,2,3,4,5 D) 2,1,4,3,5 | back 127 B) 1,2,3,5,4 |
front 128 How does termination of translation take place? A) The poly-A tail is reached B) A stop codon is reached C) The 5' cap is reached D) The end of the mRNA molecule is reached | back 128 B) A stop codon is reached |
front 129 Post-translational modifications of proteins may include the ____. A) addition of carbohydrates to form a glycoprotein B) addition of a poly-A tail C) addition of a 5' cap D) removal of introns | back 129 A) addition of carbohydrates to form a glycoprotein |
front 130 During elongation, which site in the ribosome represents the location where ac codon is being read? A) E site B) A site C) P site D) the small ribosomal subunit | back 130 B) A site |
front 131 What must occur before a newly made polypeptide is secreted from a cell? A) Its signal sequence must target it to the ER, after which it goes to the Golgi. B) Its signal sequence must target it to the plasma membrane, where it causes exocytosis C) Its signal sequence must be cleaved off before the polypeptide can enter the ER D) It must be translated by a ribosome that remains free within the cytosol | back 131 A) Its signal sequence must target it to the ER, after which it goes to the Golgi. |
front 132 A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The following charged transfer RNA molecules (with their anticodons shown in the 3' to 5' direction) are available. Two of them can correctly match the mRNA so that a dipeptide can form. The dipeptide that will form will be A) cysteine-alanine. B) proline-threonine. C) glycine-cysteine. D) alanine-alanine. E) threonine-glycine. | back 132 B) proline-threonine. |
front 133 Which of the following is the first event to take place in translation in eukaryotes? A) elongation of the polypeptide B) base pairing of activated methionine-tRNA to AUG of the messenger RNA C) binding of the larger ribosomal subunit to smaller ribosomal subunits D) covalent bonding between the first two amino acids E) the small subunit of the ribosome recognizes and attaches to the 5' cap of mRNA | back 133 E) the small subunit of the ribosome recognizes and attaches to the 5' cap of mRNA |
front 134 A signal peptide ____. A) signals the initiation of transcription B) Helps target a protein to the ER C) Directs an mRNA molecule into a cisternal space of the ER D) terminates translation of messenger RNA | back 134 B) Helps target a protein to the ER |
front 135 Which of the following statements is true about protein synthesis in prokaryotes? A) Extensive RNA processing is required before prokaryotic transcripts can be translated. B) Translation can begin while transcription is still in progress. C) Prokaryotic cells have complicated mechanisms for targeting proteins to the appropriate cellular organelles. D) Translation requires antibiotic activity. E) Unlike eukaryotes, prokaryotes require no initiation or elongation factors. | back 135 B) Translation can begin while transcription is still in progress. |
front 136 The most commonly occurring mutation in people with cystic fibrosis is a deletion of a single codon. This results in A) a base-pair substitution. B) a nucleotide mismatch. C) a frameshift mutation. D) a polypeptide missing an amino acid. E) a nonsense mutation. | back 136 D) a polypeptide missing an amino acid. |
front 137 How might a single base substitution in the sequence of a gene affect the amino acid sequence of a protein encoded by the gene, and why? A) The amino acid sequence would be substantially altered, because the reading frame would change with a single base substitution. B) Only a single amino acid could change, because the reading frame is unaffected. C) All amino acids following the substitution would be affected, because the reading frame would be shifted. D) It is not possible for a single base substitution to affect protein structure, because each codon is three bases long. | back 137 B) Only a single amino acid could change, because the reading frame is unaffected. |
front 138 What do we mean when we use the terms monohybrid cross and dihybrid cross? A) A monohybrid cross results in a 9:3:3:1 ration whereas a dihybrid cross gives a 3:1 ratio. B) A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents. C) A dihybrid cross involves organisms that are heterozygous for two characters that are being studied, and a monohybrid cross involves organisms that are heterozygous for only one character being studied. D) A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations. | back 138 C) A dihybrid cross involves organisms that are heterozygous for two characters that are being studied, and a monohybrid cross involves organisms that are heterozygous for only one character being studied. |
front 139 The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following? A) All of the genes controlling the traits were located on the same chromosome B) None of the traits obeyed the law of segregation C) All of the genes controlling the traits behaved as if they were on different chromosomes D) The diploid number of chromosomes in the pea plant was 7 | back 139 C) All of the genes controlling the traits behaved as if they were on different chromosomes |
front 140 In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the F₁ offspring is allowed to self-pollinate. The predicted outcome of the F₂ is diagrammed in the Punnett square shown in Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Which of the boxes marked 1-4 correspond to plants with dark leaves? A) 1 only B) 1 and 2 C) 2 and 3 D) 4 only E) 1, 2, and 3 | back 140 E) 1, 2, and 3 |
front 141 Albinism is an autosomal (not sex-linked) recessive trait. A man and woman are both of normal pigmentation and have one child out of three who is albino (without melanin pigmentation). What are the genotypes of the albino's parents? A) Both are heterozygous B) One parent must be homozygous for the recessive allele; the other parent must be homozygous dominant, homozygous recessive, or heterozygous. C) One parent must be homozygous dominant; the other parent must be heterozygous. D) One parent must be heterozygous; the other parent can be homozygous dominant, homozygous recessive, or heterozygous. | back 141 A) Both are heterozygous |
front 142 Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of crosses BbTt x BBtt will be expected to have black fur and long tails? A) 9/16 B) 1/2 C) 3/8 D) 1/16 | back 142 B) 1/2 |
front 143 A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative. Which of the following is a possible phenotype for the father? A) A negative B) O negative C) B positive D) AB negative E) impossible to determine | back 143 C) B positive |
front 144 In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three allele for this autosomal gene: IA, IB, and i. The IA allele codes for the A glycoprotein. The IB allele codes for the B glycoprotein, and the i allele doesn't code for any membrane glycoprotein. IA and IB are codominant, and i is recessive to both IA and IB. People with type A blood have the genotypes IAIA or IAi, people with type B blood are IBIB or IBi, people with type AB blood are IAIB, and people with type O blood are ii. If a woman with type AB blood marries a man with type O blood, which of the following blood types could their children possibly have? A) AB and O B) A and B C) A, B, AB, and O D) A, B, and O | back 144 B) A and B |
front 145 The following question refer to the pedigree chart in Figure 14.2 for
a family, some of whose members exhibit the dominant trait, W.
Affected individuals are indicated by a dark square or circle. A) WW B) Ww C) ww D) WW or ww E) ww or Ww | back 145 C) ww |
front 146 What is the likelihood that the progeny of IV-3 and IV-4 will have the trait? A) 0% B) 25% C) 50% D) 75% E) 100% | back 146 C) 50% |
front 147 What is the probability that individual III-1 is Ww? A) 3/4 B) 1/4 C) 2/4 D) 2/3 E) 1 | back 147 E) 1 |
front 148 Use the following pedigree (Figure 14.3) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis. From this pedigree, how does this trait seem to be inherited? A) from mothers B) as an autosomal recessive C) as a result of epistasis D) as an autosomal dominant E) as an incomplete dominant | back 148 D) as an autosomal dominant |
front 149 Which of the following statements is a correct explanation for the observation that all offspring exhibit a phenotype for a particular trait that appears to be a blend of the two parental varieties? A) Neither of the parental genes is dominant over the other B) The genes are linked and do not separate during meiosis C) The genes for the trait are recessive in both of the parents D) The genes for the trait are dominant in both parents | back 149 A) Neither of the parental genes is dominant over the other |
front 150 A gene for the MN blood group has codominant alleles M and N. If both children are of blood type M, which of the following is possible? A) Each parent is either M or MN. B) Each parent must be type M C) Neither parent can have the N allele D) Both children are heterozygous for this gene | back 150 A) Each parent is either M or MN. |
front 151 In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short? A) 1 B) 1/2 C) 1/4 D) 1/6 E) 0 | back 151 E) 0 |
front 152 In the cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC? A) 1/4 B) 1/8 C) 1/16 D) 1/32 E) 1/64 | back 152 E) 1/64 |
front 153 A black guinea pig crossed with an albino guinea pig produced twelve black offspring. When the albino was crossed with a second black animal, six blacks and six albinos were obtained. What is the best explanation for this genetic situation? A) Albino is recessive; black is dominant B) Albino is recessie; black is codominant C) Albino and black are codominant D) Albino is dominant; black is incompletely dominant | back 153 A) Albino is recessive; black is dominant |
front 154 Gray seed color in peas is dominant to white. Assume that Mendel conducted a series of experiments where plants with gray seeds were crossed among themselves, and the following progeny were produced: 302 gray and 98 white. (i) What is the probable genotype of each parent? (ii) Based on your answer, what genotypic and phenotypic ratios are expected in these progeny? G= gray and g=white A) Gg x Gg; genotypic= 1:2:1, phenotypic = 3:1 B) gg x Gg; genotypic = 1:2, phenotypic = 3:1 C) GG x gg; genotypic = 3:1, phenotypic= 1:2:1 D) GG x Gg; genotypic = 1:2:1, phenotypic = 2:1 | back 154 A) Gg x Gg; genotypic= 1:2:1, phenotypic = 3:1 |
front 155 In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the F₁ offspring is allowed to self-pollinate. The predicted outcome of the F₂ is diagrammed in the Punnett square shown in Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Which of the boxes correspond to plants with a heterozygous genotype? A) 1 B) 1 and 2 C) 1, 2, 3 and 4 D) 2 and 3 | back 155 D) 2 and 3 |
front 156 Which of the plants will be true-breeding? A) 1 and 4 only B) 2 and 3 only C) 1, 2, 3, and 4 D) 1 only E) 1 and 2 only | back 156 A) 1 and 4 only |
front 157 Skin color in a certain species of fish is inherited via a single
gene with four different alleles. A) 1 B) 2 C) 4 D) 8 E) 16 | back 157 C) 4 |
front 158 Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division? A) prophase I of meiosis B) anaphase II of meiosis C) metaphase I of meiosis D) anaphase I of meiosis E) anaphase of mitosis | back 158 D) anaphase I of meiosis |
front 159 What was the most significant conclusion that Gregor Mendel drew from
his experiments with pea plants? B) Traits are inherited in discrete units, and are not the results of "blending." C) Recessive genes occur more frequently in the F₁ generation than do dominant ones. D) Genes are composed of DNA. E) An organism that is homozygous for many recessive traits is at a disadvantage. | back 159 B) Traits are inherited in discrete units, and are not the results of "blending." |
front 160 The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason? A) segregation of maternal and paternal alleles B) recurrent mutations forming new alleles C) crossing over during prophase I D) different possible assortment of chromosomes into gametes E) the tendency for dominant alleles to segregate together | back 160 C) crossing over during prophase I |
front 161 Why did Mendel continue some of his experiments to the F₂ or F₃
generation? B) to observe whether or not a recessive trait would reappear C) to observe whether or not the dominant trait would reappear D) to distinguish which alleles were segregating E) to be able to describe the frequency of recombination | back 161 B) to observe whether or not a recessive trait would reappear |
front 162 Which of the following about independent assortment and segregation is correct? A) The law of independent assortment requires describing two or more genes relative to one another. B) The law of segregation requires describing two or more genes relative to one another. C) The law of segregation requires having two or more generations to describe. D) The law of independent assortment is accounted for by observations of prophase I. E) The law of segregation is accounted for by anaphase of mitosis. | back 162 A) The law of independent assortment requires describing two or more genes relative to one another. |
front 163 A sexually reproducing animal has two unlinked genes, one for head
shape (H) and one for tail length (T). Its genotype is HhTt. Which of
the following genotypes is possible in a gamete from this organism?
B) Hh C) HhTt D) T E) HT | back 163 E) HT |
front 164 An original section of DNA has the base sequence AGCGTTACCGT. A mutation in this DNA strand results in the base sequence AGGCGTTACCGT. This change represents _____. A) A silent mutation B) A frameshift mutation C) A point mutation D) A missense mutation | back 164 B) A frameshift mutation |
front 165 A single base substitution mutation is least likely to be deleterious when the base change results in ____. A) A codon that specifies the same amino acid as the original codon B) A stop codon C) An amino acid substitution that alters the tertiary structure of the protein D) An amino acid substitution at the active site of an enzyme | back 165 A) A codon that specifies the same amino acid as the original codon |
front 166 Rank the following one-base point mutations (from most likely to least likely) with respect to their likelihood of affecting the structure of the corresponding polypeptide. 1. insertion mutation deep within an intron 2. substitution mutation at the third position of an exonic codon 3. substitution mutation at the second position of an exonic codon 4. deletion mutation within the first exon of the gene A) 2,1,4,3 B) 1,2,3,4 C) 3,1,4,2 D) 4,3,2,1 | back 166 D) 4,3,2,1 |
front 167 Of the following, which is the most current description of a gene? A) A unit of heredity that causes formation of a phenotypic characteristic B) A DNA subunit that codes for a single complete protein C) a discrete unit of hereditary information that consists of a sequence of amino acids D) A DNA sequence that is expressed to form a functional product: either RNA or polypeptide | back 167 D) A DNA sequence that is expressed to form a functional product: either RNA or polypeptide |
front 168 Which of the following types of mutation, resulting in an error in the mRNA just after the AUG start of translation, is likely to have the most serious effect on the polypeptide product? A) a substitution of the first nucleotide of the GGG codon B) a substitution of the third nucleotide in an ACC codon C) a deletion of two nucleotides D) a deletion of a codon | back 168 C) a deletion of two nucleotides |
front 169 A nonsense mutation in a gene ____. A) has no effect not he amino acid sequence of the encoded protein B) introduces a premature stop codon into the mRNA C) alters the reading frame of the mRNA D) changes an amino acid in the encoded protein | back 169 B) introduces a premature stop codon into the mRNA |
front 170 Which of the following DNA mutations is most likely to damage the protein it specifies? A) a codon deletion B) an addition of three nucleotides C) a base-pair deletion D) a substitution in the last base of a codon | back 170 C) a base-pair deletion |