UTC BIO UNIT 2 TEST STUDY SET

E

CH.10 A cell in the G2 phase: A. has visibly distinct chromosomes. B. lacks a visible nuclear membrane. C. is in mitosis. D. is in cytokinesis. E. has twice the amount of DNA present in a telophase nucleus.

E

CH.10 An animal with a diploid number of 36 chromosomes will have ____ chromosomes in its gametes and ____ chromosomes in its somatic cells. A. 36; 36 B. 36; 18 C. 18; 18 D. 36; 72 E. 18; 36

D

CH.10 Gametophyte plants produce gametes using: A. meiosis. B. sporogenesis. C. fertilization. D. mitosis. E. polyploidy.

A

CH.10 Homologous chromosomes can be identified and/or characterized by: A. their similar and characteristic staining patterns. B. those chromosomes that do not have partners. C. the different centromere positions in the maternal vs. paternal chromosomes. D. maternal chromosomes only. E. paternal chromosomes only.

C

CH.10 In a human cell at prophase I, there are ____ tetrads. A. 92 B. 46 C. 23 D. 2 E. 4 .

A

CH.10 In a prometaphase cell, the: A. nucleolus disappears. B. duplicated chromosomes become visible with the light microscope. C. spindle fibers pull chromosomes to opposite sides of the cell. D. mitotic spindle is beginning to assemble. E. nuclear envelope is visible with the light microscope.

B

CH.10 When is a cell in metaphase? A. The chromosomes are visible as threadlike structures. B. The chromosomes are aligned at the midplane of the cell. C. The chromosome are separated into distinct groups at opposite poles of the cell. D. The nuclear envelope is clearly visible. E. Cytokinesis is occurring.

E

CH.10 Which of the following is produced by meiosis? A. somatic cells B. polyploid cells C. animal gametes D. diploid cells E. zygotes

E

CH.10 https://cnow.apps.ng.cengage.com/ilrn/books/smbp10r/smbp10r.10.Figure_10-2.1.png The process occurring at arrow 3 in the accompanying figure is: A. mitosis. B. G1. C. fusion. D. fertilization. E. meiosis.

C

CH.11 What are the possible genotypes of a female child from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? A. XHYH B. HH C. X^H X^H or X^H X^h D. Hh E. XHXh

E

CH.11 What are the possible phenotypes of the children if the mother has Type O blood and the father has type AB blood? (Use the Punnett square to verify your answer.) A. A, B, O B. AB only C. A, AB D. A, B E. A, B, AB

B

CH.12 Errors in DNA replication can come from: A. DNA polymerase. B. the sun's UV radiation. C. complementary base pairing. D. DNA ligase. E. Okazaki fragments.

A

CH.12 If DNA replication rejoined the two parental strands, it would be termed: A. conservative. B. parental. C. dispersive. D. gradient. E. semiconservative.

A

CH.12 In order to fit the X-ray crystallographic data, the two DNA strands must be ____ to each other. A. antiparallel B. complementary C. uncomplementary D. parallel E. semiconservative

E

CH.12 In replication, once the DNA strands have been separated, reformation of the double helix is prevented by: A. DNA helicase enzyme. B. ATP. C. DNA primase. D. DNA polymerases. E. single-strand binding proteins.

E

CH.12 Okazaki fragments are joined together by: A. DNA polymerase. B. RNA ligase. C. RNA polymerase. D. primase. E. DNA ligase.

E

CH.12 The chemical group box 1 in the accompanying figure is a: A. purine. B. amino acid. C. phosphate. D. covalent bond. E. pyrimidine.

E

CH.12 The ends of eukaryotic chromosomes can be lengthened by: A. DNA polymerase. B. mismatch repair enzymes. C. primase. D. apoptosis. E. telomerase.

C

CH.12 Which of the following adds new nucleotides to a growing DNA chain? A. primase B. RNA polymerase C. DNA polymerase D. DNA helicase E. RNA primer

B

CH.12 X-ray diffraction studies are used to determine the: A. wavelength of X-rays. B. distances between atoms of molecules. C. sequence of nucleotides in nucleic acid molecules. D. identity of an unknown chemical. E. sequence of amino acids in protein molecules

D

CH.13 Introns in pre-mRNA: A. move within the mRNA, giving rise to new exon combinations. B. protect pre-mRNA from enzyme degradation. C. code for specific protein domains. D. are spliced out of the message. E. code for important amino acid sequences.

A

CH.13 Why is only one strand of DNA transcribed into mRNA? A. Because transcribing both DNA strands would produce different amino acid sequences B. Because the other strand is transcribed directly into amino acids C. Because all genes are located on the same DNA strand, while the other strand acts as protection D. Because mRNA is only required in small quantities E. Because the other strand would produce the same amino acid sequence in reverse order

E

CH.13 https://cnow.apps.ng.cengage.com/ilrn/books/smbp10r/smbp10r.13.Figure_13-3.1.png Refer to the accompanying figure. Which label indicates the location of the promoter region? A. A B. B C. C D. D E. E

C

CH.11 Color-blindness is more common in males than in females because: A. color-blindness is an autosomal trait. B. females need to inherit only one copy of the recessive color blindness allele to express the trait. C. males only need to inherit the recessive maternal color blindness allele to be color blind. D. all females are hemizygous for the color blindness trait. E. males can only inherit paternal color blindness alleles.

B

CH.11 If a couple is planning on having two children, what is the probability that both will be male? A. 0 B. 1/4 C. 1/2 D. 3/4 E. 1

B

CH.11 If an allele is dominant, then: A. its phenotype is always expressed in all generations. B. one allele can mask the expression of a recessive allele in a hybrid. C. its phenotype is only expressed in heterozygous individuals. D. its phenotype is more beneficial than the recessive phenotype. E. its phenotype is expressed only in homozygous individuals..

B

CH.10 A cell is dividing by binary fission. What can you conclude? A. Mitosis has taken place without cytokinesis. B. The cell is prokaryotic. C. The cell cycle is out of control. D. Homologous chromosomes have already paired. E. The cyclin-Cdk complex is no longer phosphorylating enzymes.

B

CH.10 At the completion of oogenesis, ____ are produced. A. haploid spores B. one egg cell and three polar bodies C. four sperm cells D. male gametes E. four egg cells

D

CH.10 Check My Work Chromosomes are duplicated during the ____ of the cell cycle. A. prophase B. G1 phase C. metaphase D. S phase E. G2 phase

B

CH.10 Cytokinesis in animal cells involves contraction of a ring of ____ microfilaments. A. cyclin plus actin B. actin plus myosin C. cohesin plus actin D. tubulin plus actin E. cyclin plus myosin

D

CH.10 Cytokinesis in plant cells occurs via the formation of a(n): A. Golgi complex. B. aster. C. cell wall. D. cell plate. E. mitotic spindle.

B

CH.10 Duplicated centrioles move to opposite poles of a dividing ____ cell during the ____ of the cell cycle. A. animal; interphase B. animal; prophase C. plant; anaphase D. plant; metaphase E. prokaryotic; metaphase

E

CH.10 During prophase I, each chiasma represents: A. the remnant of the nucleolus. B. a newly formed haploid gamete. C. the site where sister chromatids are connected. D. the remnants of the nuclear envelope. E. a site of crossing-over.

E

CH.10 During prophase, ____ is(are) compacted into visible chromosomes. A. centrioles B. tetrads C. centromeres D. kinetochores E. chromatin

E

CH.10 Eukaryotic chromosomes are composed of: A. proteins only. B. karyotypes. C. circular chromatin. D. DNA only. E. chromatin.

B

CH.10 In unwound chromatin, nucleosomes are organized into large coiled loops held together by: A. kinetochore proteins. B. scaffolding proteins. C. histones. D. condensins. E. centromeres.

D

CH.10 Nucleosomes are best described as: A. eukaryotic DNA associated with scaffolding proteins. B. prokaryotic DNA associated with histone proteins. C. eukaryotic DNA associated with nonhistone proteins. D. eukaryotic DNA associated with histone proteins. E. prokaryotic DNA associated with nonhistone proteins.

B

CH.10 Plant hormones known as ____ stimulate mitosis. A. scaffolding proteins B. cytokinins C. cohesins D. growth factors E. cyclins

C

CH.10 The ____ is responsible for the separation of the chromosomes during the ____ of mitosis. A. centromere; telophase B. mitotic spindle; interphase C. mitotic spindle; anaphase D. kinetochore; prophase E. cell wall; anaphase

E

CH.10 The cell cycle of a typical somatic cell consists of the ____ and M phases. A. mitosis B. meiosis II C. meiosis I D. crossing-over E. interphase

C

CH.10 The function of nucleosomes is to: A. prevent RNA from tangling with DNA during transcription. B. make RNA synthesis possible. C. prevent DNA strands from tangling. D. help DNA replicate. E. prevent histones from tangling.

D

CH.10 To control the cell cycle: A. M-Cdk inhibits mitosis. B. Cdks are active only when they are released from cyclins. C. the anaphase-promoting complex stimulates DNA replication. D. cyclins fluctuate during the cell cycle. E. the activity of Cdks stays the same throughout the cell cycle.

C

CH.10 To prevent disastrous consequences, the eukaryotic cell cycle is controlled by: A. environmental signals. B. replication. C. a series of cell cycle checkpoints. D. a very detailed, rigid genetic program. E. the mitochondria.

C

CH.10 Which event occurs in prophase II? A. disappearance of the spindle B. formation of the chiasmata C. condensation of the chromatin into chromosomes D. crossing over occurs. E. formation of the cleavage furrow

B

CH.10 ____ contain identical DNA sequences and are held together by ____ during mitosis. A. Daughter chromosomes; ionic bonding B. Sister chromatids; cohesin proteins C. Daughter chromosomes; hydrogen bonding D. Sister chromosomes; histone proteins E. Sister chromatids; spindle fibers

E

CH.10 ____ organized on chromosomes carry the information that controls the functions of the cell. A. Nucleosomes B. Asters C. Proteins D. Histones E. Genes

C

CH.10 https://cnow.apps.ng.cengage.com/ilrn/books/smbp10r/smbp10r.10.Figure_10-1.1.png In the accompanying figure, which combination of letters accurately represents two homologous chromosomes? A. A and B B. A and C C. A and D D. B and F E. D and E

B

CH.11 A brown-eyed couple already has a child with blue eyes. What is the probability that their next child will have blue eyes, assuming that the brown eye allele is dominant and the blue eye allele is recessive? A. 0 B. 1/4 C. 1/2 D. 3/4 E. 1

C

CH.11 A brown-eyed couple heterozygous for eye color are planning on having two children. What is the probability that both children will have blue eyes, assuming brown eyes is dominant and blue eyes is recessive? A. 0 B. 1/32 C. 1/16 D. 1/4 E. 1/2

C

CH.11 A couple has already had three girls with cystic fibrosis, and are hoping to have a normal child for their fourth. What are the chances that the fourth child will be a normal male? A. 1/8 B. 1/4 C. 3/8 D. 1/2 E. 0

B

CH.11 A gene that affects, prevents, or masks the expression of a gene at another locus is a(n) ____ gene. A. recessive B. epistatic C. codominant D. pleiotropic E. dominant

B

CH.11 A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb). If there are 800 F1 progeny, the expected phenotypes would be: A. 800 normal gray B. 200 normal gray, 200 normal black, 200 vestigial gray, and 200 vestigial black C. 800 vestigial black D. 450 normal gray, 150 normal black, 150 vestigial gray, and 50 vestigial black E. 400 normal gray and 400 vestigial black

C

CH.11 A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb). Out of 800 total F1 progeny, 296 are normal gray, 328 are vestigial black, 99 are normal black, and 77 are vestigial gray. What is the best explanation for this result? A. The loci for wing length and body color are on sex chromosomes. B. The loci for wing length and body color are on different chromosomes. C. The loci for wing length and body color are on the same chromosome. D. Mendel's principle of independent assortment accounts for this. E. Mendel's principle of segregation accounts for this.

C

CH.11 A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb). This type of cross is known as a: A. monohybrid cross. B. chromosomal cross. C. dihybrid cross D. polytropic cross. E. true breeding cross.

A

CH.11 All calico cats are female because: A. the calico phenotype is caused by random X chromosome inactivation. B. one X chromosome and one Y chromosome are needed for the expression of the calico phenotype. C. two recessive X-linked genes are required for the calico phenotype. D. Y-linked genes prevent the expression of the calico phenotype. E. the calico phenotype is a Y-linked trait. Hide Feedback

C

CH.11 An organism with the genotype of AaXx can produce gametes containing ____ if the two genes are unlinked. A. AX or ax B. AaXx only C. either AX, Ax, aX, ax D. either Aa or Xx E. A,a,X,x

B

CH.11 Breeding a yellow dog with a brown dog produced puppies with both yellow and brown hairs intermixed. This is an example of: A. incomplete dominance. B. codominance. C. a polygenic trait. D. variegation. E. epistasis.

E

CH.13 Frameshift mutations result from the: A. substitution of a stop codon for an amino acid-specifying codon. B. substitution of a start codon for an amino acid codon. C. substitution of one base pair for another. D. substitution of more than one base pair. E. insertion or deletion of one or two base pairs.

C

CH.13 Garrod first proposed that: A. mutations were heritable. B. metabolic defects did not occur in humans. C. metabolic defects were due to the lack of an enzyme. D. metabolic defects were due to chromosomal changes. E. metabolic defects were due to excess enzyme production.

E

CH.13 How does the first nucleotide at the 5¢ end of a new mRNA chain differ from the other nucleotides in the chain? A. The first nucleotide is always a cytosine. B. The first nucleotide is always a modified cytosine. C. The first nucleotide does not retain its triphosphate group, while the others in the chain do. D. The first nucleotide is always a uracil. E. The first nucleotide retains its triphosphate group, while the others do not.

C

CH.13 If an mRNA transcript of the human XYZ gene was introduced into a yeast cell, the transcript would be: A. degraded immediately. B. transcribed into the XYZ mRNA. C. translated into the human XYZ protein. D. translated into a yeast-specific gene. E. exported outside the cell.

B

CH.13 In the Beadle and Tatum study, it was important that Neurospora was haploid because: A. haploid molds have simpler nutritional requirements than do diploid molds. B. a mutation that arises is not masked by a normal allele on a homologous chromosome. C. haploid Neurospora will always mutate. D. diploid Neurospora will always mutate. E. it is easier to grow haploid molds in the laboratory.

B

CH.13 Linus Pauling demonstrated that: A. the structure of hemoglobin was altered by mutations in any of a dozen genes. B. mutations alter the structure of RNA, but not proteins. C. the structure of hemoglobin was altered by a mutation of a single gene. D. mutations were inherited. E. mutations only caused defects in enzymes.

D

CH.13 Mobile genetic elements, such as transposons and retrotransposons, are significant because: A. transposons are the only mechanisms of genetic diversity. B. only corn contain transposons. C. they are present in the human genome, but not in the genomes of other organisms. D. mutations caused by retrotransposons contribute greatly to a species' evolution. E. the entire human genome is composed of transposons.

E

Ch.14 In Jacob and Monad's analysis of mutant E. coli strains in their study of the lac operon, the absence of the β-galactosidase enzyme, but presence of the lactose permease and galactosidase transacetylase enzymes, indicated that the mutation occurred in the _____ of the lac operon. A. lacA gene B. regulatory region C. lacY gene D. heterochromatin E. lacZ gene

C

Ch.14 In an inducible operon, the ____ is the molecular switch that controls gene expression, while the ____ is the DNA region where RNA polymerase binds to begin transcription. A. promoter; operator B. repressor; inducer C. operator; promoter D. lactose; promoter E. inducer; repressor

E

Ch.14 In female mammals, most of the inactive X chromosome becomes: A. associated with histones. B. replicated. C. transcriptionally active. D. euchromatin. E. heterochromatin.

D

Ch.14 Lactose induces the transcription of the lactose operon by: A. stimulating lactose metabolism in the cell. B. binding to the allosteric site of RNA polymerase. C. inhibiting the activity of CAP. D. binding to the allosteric site of the repressor after being converted to allolactose. E. first being converted to glucose, then binding to the repressor.

B

Ch.14 Leucine zipper proteins regulate transcription by: A. forming dimers with zinc finger transcription factors. B. binding to and activating DNA regulatory elements such as enhancers and silencers. C. degrading RNA polymerase. D. changing the chromatin structure to heterochromatin E. blocking DNA regulatory elements.

B

Ch.14 Long coding RNAs (lnc RNAs) are characterized by: A. the lack of a polyA tail. B. their roles in regulating transcription. C. binding to zinc finger transcription factors. D. being less than 100 bases in length. E. the presence of introns.

E

Ch.14 Many genes encode proteins that are always needed by a cell, and therefore are always expressed. Such genes are known as: A. repressible. B. operons. C. inducible. D. promoter. E. constitutive.

B

Ch.14 Methylated DNA sequences typically contain ____ genes. A. constitutive B. transcriptionally inactive C. alternatively spliced D. housekeeping E. transcriptionally active

C

Ch.14 Proteolytic processing of eukaryotic proteins involves: A. the addition of one or more functional groups to a precursor protein. B. removal of histones. C. removal of a portion of the polypeptide chain from an inactive protein precursor. D. removing the cap and tail of a newly synthesized protein. E. the removal of one or more functional groups from a precursor protein.

E

Ch.14 Repressible genes are usually actively transcribed when: A. quantities of precursor materials are high. B. there is no other substrate that can be used by the cell. C. tryptophan accumulates in the cell. D. repressor molecules bind to the promoter. E. the end product of the enzymes encoded by these genes is low.

D

Ch.14 The function of the catabolite activator protein (CAP) is to: A. decrease RNA polymerase activity. B. deactivate RNA polymerase after binding to AMP. C. bind to and activate RNA polymerase directly. D. increase the affinity of the promoter region for RNA polymerase. E. block RNA polymerase from accessing the DNA template.

E

Ch.14 The gene that codes for the repressor protein of the E. coli lac operon is constitutive; therefore, the lac operon is typically: A. turned on. B. unaffected by the repressor protein. C. activated by the repressor protein.. D. constitutive. E. turned off.

D

Ch.14 The organization of the lac operon, from upstream to downstream sequences, is represented by: A. operator -> promoter -> lacZ -> lacY -> lac A B. repressor gene -> lacZ -> lacY -> lac A C. lacZ -> lacY -> lac A -> operator -> promoter D. promoter -> operator -> lacZ -> lacY -> lac A E. lacZ -> lacY -> lac A

A

Ch.14 Transcription factors that increase the rate of transcription by binding a target in the general transcriptional machinery and an enhancer are known specifically as: A. activator proteins. B. leucine zipper proteins. C. lnc RNAs D. helix-turn-helix proteins. E. zinc finger proteins.

A

Ch.14 Transcriptional activation of the lac operon requires both high lactose and low glucose conditions because: A. RNA polymerase requires CAP in order to bind to the lac operon promoter, even in the absence of the repressor. B. glucose and lactose compete for binding to, and inactivation of, the lac operon repressor. C. high glucose concentrations stimulate CAP activity. D. glucose normally activates the lac operon repressor. E. RNA polymerase is inactive in the presence of high glucose concentrations.

B

Ch.14 Translational controls regulate the: A. activity of a protein product. B. uptake of nucleic acids into the cell. C. rate at which an mRNA molecule is translated. D. rate at which an mRNA molecule is synthesized. E. attachment of phosphate groups to polypeptides.

A

Ch.14 _____ are densely staining, highly compacted, and transcriptionally inactive regions of chromatin. A. Heterochromatin B. Histone-dependent chromatin C. Homochromatin D. Primary chromatin E. Euchromatin

B

Ch.14 cAMP levels decrease when __________. This results in __________ of CAP. A. glucose levels decrease; activation B. glucose levels increase; inhibition C. lactose levels decrease; inhibition D. lactose levels increase; activation E. glucose levels decrease; inhibition

E

Ch.14 https://cnow.apps.ng.cengage.com/ilrn/books/smbp10r/smbp10r.14.Figure_14-1.1.png Refer to the accompanying figure. The area of the tryptophan operon labeled as 3 is the: A. ribosome. B. RNA polymerase. C. promoter. D. repressor gene. E. operator.

E

Ch.14 https://cnow.apps.ng.cengage.com/ilrn/books/smbp10r/smbp10r.14.Figure_14-1.1.png The promoter in the accompanying figure is labeled as ____. A. 1 B. 6 C. 3 D. 7 E. 2

D

CH.11 In humans, assume that the allele for brown eyes is dominant and the allele for blue eyes is recessive. If two brown-eyed individuals have a child with blue eyes, that means: A. both parents are homozygous for brown eyes. B. there is a 1/4 chance that their second child will have brown eyes. C. there is a 1/2 chance that their second child will have blue eyes. D. both parents are heterozygous for eye color. E. there is a 3/4 chance that their second child will have blue eyes

A

CH.11 In peas, Mendel found that tall plants and yellow peas are dominant. The phenotype for a pea plant with the genotype TTyy would be: A. tall with green peas. B. tall with yellow peas. C. short with green peas. D. intermediate height with greenish-yellow peas. E. short with yellow peas.

E

CH.11 In rabbits, coat color is governed by four alleles: C for dark gray, Cch for chinchilla, Ch for himalayan, and c for white. This is an example of: A. X-linked genes. B. codominance. C. incomplete dominance. D. epistasis. E. multiple alleles.

A

CH.11 Mating a true-breeding pink rose plant with a true-breeding pink rose plant will produce: A. plants with only pink roses. B. plants with pink, red, and white roses. C. plants with white or red roses in a 3:1 ratio. D. plants with red or white roses in a 3:1 ratio. E. plants with only red roses.

B

CH.11 Mendel's principle of independent assortment is explained by which metaphase I event? A. separation of homologous chromosomes B. random orientation of homologous chromosomes on the metaphase plate C. crossing-over D. synapsis E. cytokinesis

B

CH.11 Mendel's principle of segregation states that: A. dominant and recessive alleles segregate during meiosis. B. alleles separate from each other before forming gametes. C. true-breeding parents produce offspring of the same phenotype. D. hybrids will express a phenotype intermediate between the two parental phenotypes.

A

CH.11 The chromosome theory of inheritance, developed by Sutton and Boveri, provided an explanation for Mendel's principle of segregation by proposing that: A. homologous chromosomes segregate during meiosis. B. all genes are linked on a single chromosome. C. linked genes are inherited together. D. sister chromatids separate during mitosis. E. homologous chromosomes segregate during mitosis.

E

CH.11 The height of pea plants from a cross between parent plants heterozygous for height, in which tall is dominant, would be: A. all short. B. all tall. C. 1 tall : 3 short. D. 2 short : 2 tall. E. 3 tall : 1 short.

C

CH.11 The offspring of two heterozygous gray-bodied, normal-winged flies should be 50% gray-bodied/normal wings (BbRr) and 50% black-bodied/vestigial wings (bbrr) because these alleles are linked. Suppose a small number, say 15%, of the offspring are instead black-bodied with normal wings. This is most likely the result of: A. mutation. B. incomplete dominance. C. crossing-over. D. an error in meiosis. E. codominance.

B

CH.11 The sex of most mammals, birds, and insects is determined by: A. the temperature. B. sex chromosomes. C. chance. D. the external environment. E. the internal environment.

B

CH.11 Two Martians fall in love and marry. One Martian is homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the probability that they will have a child with purple eyes? A. 1/2 B. 0 C. 1/4 D. 3/4 E. 1/1

D

CH.11 What are the possible phenotypes of the children if the mother's genotype is IAi for blood type and the father is IBi? (Use the Punnett square to verify your answer.) A. A, AB B. A, B C. A, B, O D. A, B, AB, O E. all AB

C

CH.11 What are the predicted phenotypes of the male children from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting characteristics? (Use the Punnett square to verify your answer.) A. 3 normal : 1 hemophilia B. 1 hemophilia : 3 normal C. 1 hemophilia : 1 normal D. all hemophiliacs E. all normal

B

CH.11 Which of the following represents the possible genotype(s) resulting from a cross between an individual homozygous for black hair (BB) and an individual homozygous for blonde hair (bb)? A. BB only B. Bb only C. bb only D. BB, Bb, and bb E. BB and Bb

E

CH.11 ____ refers to multiple independent pairs of genes having similar and additive effects on the same characteristic. A. Codominance B. Additive dominance C. Epistasis D. Complete dominance E. Polygenic inheritance

C

CH.12 A replication fork is: A. created by the action of the enzyme RNA polymerase. B. a site where one DNA strand serves as a template, but the other strand is not replicated. C. a Y-shaped structure where both DNA strands are replicated simultaneously. D. observed only in nondividing cells.. E. observed only in prokaryotic chromosomes.

B

CH.12 DNA: A. contains the sugar ribose. B. is composed of nucleotide building blocks. C. is single-stranded. D. contains pyrimidines and purines in a 2:1 ratio. E. contains uracil instead of thymine.

C

CH.12 Enzymes called ____ form breaks in the DNA molecules to prevent the formation of knots in the DNA helix during replication. A. RNA polymerases B. DNA polymerases C. topoisomerases D. DNA ligases E. single-strand binding proteins

D

CH.13 Aminoacyl-tRNA synthetases link ____ to their respective tRNA molecules by ____ bonds. A. amino acids; ionic B. amino acids; covalent C. amino acids; hydrogen D. codons; covalent E. anticodons; hydrogen

A

CH.13 Beadle and Tatum began their studies with wild-type Neurospora. How is this organism best characterized? A. A normal phenotype that will grow on minimal medium B. A virulent strain C. A strain that only grows in the wild D. A strain that will not grow in the lab E. A mutant strain that will only grow in the lab on complete medium

A

CH.13 Binding of the appropriate aminoacyl-tRNA to the A site requires: A. energy supplied by GTP. B. activation of the A site. C. no energy input. D. phosphorylation of the tRNA molecule. E. energy supplied by two ATP molecules.

E

CH.13 During protein synthesis, ribosomes: A. attach to the DNA molecule and travel along its length to produce an mRNA molecule. B. transcribe DNA into mRNA. C. translate mRNA into tRNA. D. translate tRNA into protein. E. attach to the mRNA molecule and travel along its length to produce a polypeptide.

D

CH.13 Following peptide bond formation between the amino acid in the A site on the ribosome and the growing polypeptide chain, the tRNA in the A site: A. forms a covalent bond with the P site of the ribosome. B. releases the growing polypeptide chain. C. picks up another amino acid to add to the chain. D. moves to the P site of the ribosome. E. forms a peptide bond with A site of the ribosome.

A

CH.13 A sequence of bases located upstream from a reference point occurs towards the: A. 5¢ end of the mRNA sequence. B. carboxyl end of the amino acid sequence. C. 3¢ end of the amino acid sequence. D. 3¢ end of the mRNA sequence. E. 5¢ end of the transcribed DNA strand.

B

CH.12 How are eukaryotic chromosomes replicated? A. The circular DNA molecules are replicated from multiple origins of replication bidirectionally. B. The linear DNA molecules are replicated from multiple origins of replication bidirectionally. C. The circular DNA molecules are replicated from one origin of replication bidirectionally. D. The linear DNA molecules are replicated from one origin of replication bidirectionally. E. The linear DNA molecules are replicated from one origin of replication unidirectionally.

D

CH.12 How is a bacterial chromosome replicated? A. The linear DNA molecule is replicated from one origin of replication bidirectionally. B. The circular DNA molecule is replicated from multiple origins of replication bidirectionally. C. The circular DNA molecule is replicated from one origin of replication unidirectionally. D. The circular DNA molecule is replicated from one origin of replication bidirectionally. E. The linear DNA molecule is replicated from multiple origins of replication bidirectionally.

D

CH.12 In DNA replication, the lagging strand: A. is not synthesized until the synthesis of the leading strand is completed. B. pairs with the leading strand by complementary base pairing. C. is synthesized as a complementary copy of the leading strand. D. is synthesized as a series of Okazaki fragments. E. is made up entirely of RNA primers.

E

CH.12 In the experiments of Griffith, the conversion of nonlethal R-strain bacteria to lethal S-strain bacteria: A. could not be reproduced by other researchers. B. was similar to experiments performed by Watson and Crick. C. proved that proteins are the genetic material. D. was due to genetic mutation. E. was due to transformation.

A

CH.12 Individuals with mutations in excision repair enzymes may suffer from ____ due to unrepaired DNA damage caused by ____. A. skin cancer; the sun's UV rays B. skin cancer; complementary base pairing C. colon cancer; the passing of DNA mutations to daughter cells D. skin cancer; mutations inherited from their parents E. prostate cancer; telomerase

A

CH.12 The information encoded by the DNA is specified by the: A. nucleotide sequence of the DNA molecule. B. size of a particular chromosome. C. number of bases in a DNA molecule. D. number of separate DNA strands. E. sugar-phosphate backbone of the DNA molecule.

E

CH.12 The two molecules that alternate to form the backbone of a polynucleotide chain are: A. base and sugar. B. adenine and thymine. C. base and phosphate. D. cytosine and guanine. E. sugar and phosphate.

C

CH.12 Unlike normal cells, cancerous cells: A. divide only a few times before succumbing to apoptosis. B. have reduced levels of telomerase. C. can maintain telomere length as they divide. D. have unusually short telomeres. E. lack telomeres.

D

CH.12 What happens during nucleotide excision repair? A. A mismatch mutation is repaired. B. DNA polymerase joins the repaired DNA together. C. DNA ligase adds new nucleotides to the repaired DNA strand. D. A nuclease removes the damaged DNA. E. DNA is damaged.

B

CH.12 When a mutation occurs during DNA replication, ____ replaces the incorrect nucleotide with the correct one. A. DNA ligase B. DNA polymerase C. helicase D. telomerase E. mismatch repair enzymes

E

CH.12 Which of the following statements concerning DNA is FALSE? A. DNA is a polymer of nucleotides. B. Purines and pyrimidines are complementary. C. The two chains of DNA are antiparallel. D. The structure of DNA can be described as a double helix. E. The sugar present in DNA is ribose.

C

CH.12 Which of the following statements is FALSE with regard to DNA replication? A. The position of the replication fork is constantly moving. B. DNA synthesis proceeds in the 5'->3' direction. C. The strand being copied is read in the 5'->3' direction. D. Two identical DNA polymerase molecules catalyze replication. E. Both strands replicate at the same time.

A

CH.12 Who first confirmed that the replication of DNA was semiconservative? A. Meselson and Stahl B. Avery and Griffith C. Chargaff and Hershey D. Watson and Crick E. Watson, Crick, and Wilkins

D

CH.12 Why does DNA synthesis proceed in a 5¢ to 3¢ direction? A. The 3¢ end of the polynucleotide molecule contains more phosphates than the 5¢ end. B. DNA strands are parallel to each other. C. Chromosomes are aligned in the 5¢ to 3¢ direction in the nucleus. D. DNA polymerases can only add nucleotides to the 3¢ end of a polynucleotide strand. E. DNA unzips in the 5' to 3' direction.

B

CH.12 Why is DNA able to store large amounts of information? A. Its bases can be altered from purines to pyrimidines. B. Its nucleotides can be arranged in a large number of possible sequences. C. Its sugars and phosphates can be arranged in many different sequences. D. It is composed of 20 different nucleotides. E. It is capable of assuming a wide variety of shapes.

B

CH.12 X-ray diffraction images produced by ____ were used by Watson and Crick to infer the structure of DNA.

B

CH.12 ____, the ends of eukaryotic chromosomes, shorten with every cell division. A. Kinetochores B. Telomeres C. Primosomes D. Nucleosomes E. Centromeres

B

CH.12 _____ bonds link the sugar and phosphate groups in the backbones of DNA molecules. A. Hydrogen B. Covalent phosphodiester C. Ionic D. Weak covalent E. Weak

C

CH.13 A "5¢ cap": A. prevents translation. B. is added to mRNA but not pre-mRNA. C. protects mRNA from degradation. D. prevents the binding of ribosomes. E. marks the mRNA for degradation.

B

CH.13 A ____ mutation results in the conversion of a codon specifying an amino acid to a termination codon. A. frameshift B. nonsense C. missense D. chromosomal E. silent

C

CH.13 A polyribosome is: A. one ribosome complexed to several mRNAs. B. an elongation complex in eukaryotes. C. a complex composed of several ribosomes and plus an mRNA transcript. D. an initiation complex in eukaryotes. E. a complex componsed of several ribosomes in eukaryotes.

E

CH.13 Neurospora is an ideal organism to study the effects of genetic mutations because of several reasons. What is one of these? A. Neurospora is a small mammal. B. Neurospora grows as a diploid organism. C. Neurospora contains homologous chromosomes that are easily viewed with a light microscope. D. Neurospora requires a medium supplemented with only a few nutrients that it cannot manufacture on its own. E. Neurospora mutant strains that cannot make a particular amino acid can still grow if that amino acid is added to the growth medium.

B

CH.13 One of the mRNA codons specifying the amino acid leucine is 5¢-CUA-3¢. Its corresponding anticodon is: A. 5¢-GAU-3¢. B. 3¢-GAU-5¢. C. 3¢-GAT-5¢. D. 5¢-GAT-3¢. E. 3¢-AUC-5¢.

C

CH.13 Peptidyl transferase, a _____ molecule, catalyzes the formation of _____ between amino acids at the P and A sites. A. protein; peptide bonds B. protein; hydrogen bonds C. ribozyme; peptide bonds D. protein; templates E. ribozyme; hydrogen bonds

B

CH.13 Proteins synthesized in bacteria have which of the following at their amino terminal end? A. adenine triphosphate B. N-formyl-methionine C. The AUG codon D. N-acetyl-adenine E. The UUU codon

A

CH.13 RNA synthesis is also known as: A. transcription. B. translation. C. elongation. D. reverse transcription. E. termination.

E

CH.13 The total number of different three-base combinations of the four nucleic acid bases is: A. 16. B. 20. C. 256. D. 12. E. 64

A

CH.13 These noncoding sequences located within coding regions of ____ genes are called ____. A. only eukaryotic; introns B. only eukaryotic; exons C. eukaryotic and prokaryotic; introns D. only prokaryotic; introns E. only prokaryotic; exons

D

CH.13 Translation is initiated at the ____. A. anticodon B. UAG codon C. promoter D. AUG codon E. downstream sequence

E

CH.13 What is the modern definition of a gene? A. An RNA sequence that encodes a single polypeptide B. A DNA sequence that encodes several proteins C. A DNA sequence that encodes specific amino acids D. A DNA sequence that encodes a specific polypeptide E. A DNA sequence that encodes a specific RNA or protein product

C

CH.13 Which of the following is a characteristic of uracil? A. It has the ability to bond with guanine. B. It is a purine. C. It has the ability to bond with adenine. D. It contains two nitrogenous rings. E. It has the ability to bond with cytosine.

C

CH.13 https://cnow.apps.ng.cengage.com/ilrn/books/smbp10r/smbp10r.13.Figure_13-3.1.png Refer to the accompanying figure. The component labeled as B is: A. DNase. B. reverse transcriptase. C. RNA polymerase. D. RNA primase. E. DNA polymerase

B

Ch.14 A TATA box is seen in ____ cells and is the site where ____. A. eukaryotic; DNA ligase cleaves introns B. eukaryotic; RNA polymerase binds C. prokaryotic; RNA polymerase binds D. prokaryotic; a repressor protein binds E. both prokaryotic and eukaryotic; transcription factors bind

E

Ch.14 A mutation in the IGF2 gene in pigs results in a three-fold increase in the transcriptional activity of this gene in muscle cells; therefore, these pigs are leaner and more muscular. Because this mutation is located in the noncoding region of the IGF2 gene, we can infer that: A. less IGF2 protein is synthesized in the muscle cells of these pigs. B. the mutation changes the amino acid sequence of the IGF2 protein. C. the IGF2 gene is present only in muscle cells, but not in adipose (fat) cells. D. the IGF2 gene is a prokaryotic gene. E. the mutation alters the regulatory region of the IGF2 gene.

D

Ch.14 A new operon is discovered in bacteria. Transcription of the genes in this operon is blocked by the binding of a protein to DNA sequences upstream of the operon. This is an example of ____ control. A. posttranscriptional B. inducible C. positive D. negative E. constitutive

B

Ch.14 A repressor protein would have which of the following effects on repressible genes with a negative control mechanism? A. A repressor protein does not participate in negative control mechanisms. B. Turn off transcription in the presence of a corepressor. C. Turn off transcription in the presence of an inducer. D. Stimulate transcription in the presence of a corepressor. E. Stimulate transcription in the presence of a coactivator.

A

Ch.14 An inducible operon is usually controlled by: A. an active repressor that keeps it in the "off" state. B. being turned "off," usually by the end product of the pathway. C. allolactose. D. an inducer molecule that keeps it in the "off" state. E. being active at all times.

E

Ch.14 Bacterial enzymes that are part of a rarely used catabolic pathway are usually organized into a(n) __________. A. repressible operon B. inducible zinc finger C. constitutive operon D. repressible leucine zipper E. inducible operon

C

Ch.14 Considering that virtually all cells in any given individual contain the same genetic information, how is it possible to have different cell types with unique functions? A. All cells express all proteins encoded in an organism's genes, but turn them on selectively. B. Gene expression is regulated based on the age of the organism. C. Gene expression is regulated such that different cell types produce different proteins. D. Only some cell types make protein products, while others do not. E. Cells lose unneeded genes with each cell division.

D

Ch.14 Feedback inhibition is an example of: A. a repressible system. B. transcriptional control. C. translational control. D. posttranslational control. E. constitutive control

C

Ch.14 Genes in euchromatic regions are: A. duplicated B. transcriptionally inactive. C. transcriptionally active. D. unregulated. E. repressed.

B

Ch.14 If a mutated region on chromosome 15 is inherited from the father, the child will have Prader-Willi syndrome; however, if the same mutated region is inherited from the mother, the child will have Angelman syndrome. This is an example of: A. repressible genes. B. epigenetic inheritance. C. genetic inheritance. D. eukaryotic operons. E. gene amplification.