Ch 14 17 Unit Test Genetics When Thomas

Ch 14 - 17 Unit Test Genetics

• When Thomas Hunt Morgan crossed his red-eyed F 1 generation flies to each other, the F 2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result? • A) The gene involved is on the Y chromosome. • B) The gene involved is on the X chromosome. • C) The gene involved is on an autosome, but only in males. • D) Other male-specific factors influence eye color in flies.

• B) The gene involved is on the X chromosome.

• Which of the following is the meaning of the chromosome theory of inheritance as expressed in the early twentieth century? • A) Individuals inherit particular chromosomes attached to genes. • 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.

• B) Mendelian genes are at specific loci on the chromosome and, in turn, segregate during meiosis.

• SRY is best described as ____. • A) a gene present on the X chromosome that triggers female development • 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

• C) a gene region present on the Y chromosome that triggers male development

• 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 F 1 males will have cinnabar eyes? • A) 0% • B) 25% • C) 50% • D) 100%

• Sex determination in mammals is due to the SRY gene. Which of the following could allow a person with an XX karyotype to develop a male phenotype? • A) the loss of the SRY gene from an autosome • B) translocation of SRY to a X chromosome • C) a person with an extra autosomal chromosome • D) a person with one normal and one shortened (deleted) X

• B) translocation of SRY to a X chromosome

• In humans, clear gender differentiation occurs, not at fertilization, but after the second month of gestation. What is the first event of this differentiation? • A) formation of testosterone in male embryos • B) formation of estrogens in female embryos • C) activation of SRY in male embryos and masculinization of the gonads • D) activation of SRY in females and feminization of the gonads

• B) formation of estrogens in female embryos

• All female mammals have one active X chromosome per cell instead of two. What causes this? • A) activation of the XIST gene on the X chromosome that will become the Barr body • B) activation of the BARR gene on one X chromosome, which then becomes inactive • C) inactivation of the XIST gene on the X chromosome derived from the male parent • D) attachment of methyl (CH 3) groups to the X chromosome that will remain active

• A) activation of the XIST gene on the X chromosome that will become the Barr body

• What is an adaptive advantage of recombination between linked genes? • A) Recombination is required for independent assortment. • B) Recombination must occur or genes will not assort independently. • C) New allele combinations are acted upon by natural selection. • D) The forces on the cell during meiosis II results in recombination.

• C) New allele combinations are acted upon by natural selection.

• In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in the figure above. What is the order of these genes on a chromosome map? • A) rb-cn-vg-b • B) cn-rb-b-vg • C) b-rb-cn-vg • D) vg-cn-b-rb

• c) b-rb-cn-vg

• A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12. Although his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm? • A) 1/4 will carry the two normal chromosomes, 4 and 12, 1/4 will have only the two translocation chromosomes and no normal chromosomes 4 and 12, and 1/2 will have both normal and translocated chromosomes. • B) All will carry the same translocation as the father. • C) None will carry the translocation. • D) 1/2 will be normal and the rest will have the father's translocation.

• A) 1/4 will carry the two normal chromosomes, 4 and 12, 1/4 will have only the two translocation chromosomes and no normal chromosomes 4 and 12, and 1/2 will have both normal and translocated chromosomes.

• What is a syndrome? • A) a characteristic facial appearance • B) a trait that leads to cancer at some stage in life • C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation • D) a specific characteristic that appears in conjunction with one specific aneuploidy

• C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation

• Which of the following is an example of monosomy? • A) Turner's syndrome • B) Klinefelter's syndrome • C) Down syndrome • D) trisomy X

• A) Turner's syndrome

• In his transformation experiments, what did Griffith observe? • A) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form. • B) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic. • C) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains. • D) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.

• A) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.

• 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) Alfred Hershey and Martha Chase • B) Oswald Avery, Maclyn Mc. Carty, and Colin Mac. Leod • C) Erwin Chargaff • D) Matthew Meselson and Franklin Stahl

• C) Erwin Chargaff

• 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

• A) sequence of bases

• 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 5' to 3' 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.

• B) The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.

• 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) One of the daughter cells, but not the other, would have radioactive DNA. • B) Neither of the two daughter cells would be radioactive. • C) All four bases of the DNA would be radioactive. • D) DNA in both daughter cells would be radioactive.

• D) DNA in both daughter cells would be radioactive.

• Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following? • A) the evolution of telomerase enzyme • B) DNA polymerase that cannot replicate the leading strand template to its 5' end • C) gaps left at the 5' end of the lagging strand • D) gaps left at the 3' end of the lagging strand because of the need for a primer

• C) gaps left at the 5' end of the lagging strand

• 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 3' end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5' 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

• 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

• 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

• A) relieving strain in the DNA ahead of the replication fork

• 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 DNA binding proteins

• D) single-strand DNA binding proteins

• Which of the following would you expect of a eukaryote lacking telomerase? • A) a high probability of somatic cells becoming cancerous • B) an inability to produce Okazaki fragments • C) an inability to repair thymine dimers • D) a reduction in chromosome length in gametes

• D) a reduction in chromosome length in gametes

• 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

• A) leading strands and Okazaki fragments

• DNA contains the template needed to copy itself, but it has no catalytic activity in cells. What catalyzes the formation of phosphodiester bonds between adjacent nucleotides in the DNA polymer being formed? • A) ribozymes • B) DNA polymerase • C) ATP • D) deoxyribonucleotide triphosphates

• B) DNA polymerase

• What provides the energy for the polymerization reactions in DNA synthesis? • A) ATP • B) DNA polymerase • C) breaking the hydrogen bonds between complementary DNA strands • D) the deoxyribonucleotide triphosphate substrates

• D) the deoxyribonucleotide triphosphate substrates

• What is the difference between the leading strand the lagging strand in DNA replication? • A) 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. • 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) There are different DNA polymerases involved in elongation of the leading strand 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.

• 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) nucleosome, looped domain, 30 -nm chromatin fiber • D) 30 -nm chromatin fiber, nucleosome, looped domain

• A) nucleosome, 30 -nm chromatin fiber, looped domain

• Which of the following is most critical for the association between histones and DNA? • A) Histones are small proteins. • B) Histones are highly conserved (that is, histones are very similar in every eukaryote). • C) There at least five different histone proteins in every eukaryote. • D) Histones are positively charged.

• D) Histones are positively charged.

• 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.

• More than one codon can specify the addition of the same amino acid.

• Which of the following does not occur in prokaryotic gene expression, but does occur in eukaryotic gene expression? • A) m. RNA, t. RNA, and r. RNA are transcribed. • B) RNA polymerase binds to the promoter. • C) A cap is added to the 5' end of the m. RNA. • D) RNA polymerase requires a primer to elongate the molecule.

• C) A cap is added to the 5' end of the m. RNA.

A ribozyme is ____. A) a catalyst that uses RNA as a substrate B) an RNA with catalytic activity C) an enzyme that catalyzes the association between the large and small ribosomal subunits • D) an enzyme that synthesizes RNA as part of the transcription process • •

• B) an RNA with catalytic activity

• A primary transcript in the nucleus of a eukaryotic cell is ____ the functional m. RNA, while a primary transcript in a prokaryotic cell is ____ the functional m. RNA. • 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

• B) larger than; the same size as

• A signal peptide ____. • A) directs an m. RNA molecule into the cisternal space of the ER • B) terminates translation of messenger RNA • C) helps target a protein to the ER • D) signals the initiation of transcription

• C) helps target a protein to the ER

• What type of bonding is responsible for maintaining the shape of the t. RNA molecule shown in the figure above? • A) ionic bonding between phosphates • B) hydrogen bonding between base pairs • C) van der Waals interactions between hydrogen atoms • D) peptide bonding between amino acids

• B) hydrogen bonding between base pairs

• What must occur before a newly made polypeptide is secreted from a cell? • A) It must be translated by a ribosome that remains free within the cytosol. • B) Its signal sequence must target it to the ER, after which it goes to the Golgi. • C) Its signal sequence must be cleaved off before the polypeptide can enter the ER. • D) Its signal sequence must target it to the plasma membrane, where it causes exocytosis.

• B) Its signal sequence must target it to the ER, after which it goes to the Golgi.

• Which one of the following, if missing, would usually prevent translation from starting? • A) exon • B) 5' cap • C) AUG codon • D) poly-A tail

• C) AUG codon

• How does termination of translation take place? • A) The end of the m. RNA molecule is reached. • B) A stop codon is reached. • C) The 5' cap is reached. • D) The poly-A tail is reached.

• B) A stop codon is reached.

• 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) Unlike eukaryotes, prokaryotes require no initiation or elongation factors.

• B) Translation can begin while transcription is still in progress.

• A nonsense mutation in a gene ____. • A) changes an amino acid in the encoded protein • B) has no effect on the amino acid sequence of the encoded protein • C) introduces a premature stop codon into the m. RNA • D) alters the reading frame of the m. RNA

• C) introduces a premature stop codon into the m. RNA

• 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.

• D) The resulting cells will not receive the correct number of chromosomes in the gametes, a condition known as aneuploidy.

• 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) B) C) D) n+1; n-1; n-1 n+1; n-1; n; n n+1; n-1; n-1 n+1; n; n

• A) n+1; n-1; n-1

• 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.

• C) Replication would not occur on either the leading or lagging strand.

• Hershey and Chase used a DNA-based virus for their work. What would the results have been if they had used an RNA virus? • A) With an RNA virus radioactive protein would have been in the final pellet. • B) With an RNA virus radioactive RNA would have been in the final pellet. • C) With an RNA virus neither sample would have had a radioactive pellet. • D) With an RNA virus the protein shell would have been radioactive in both samples.

• B) With an RNA virus radioactive RNA would have been in the final pellet.

• The lagging strand is characterized by a series of short segments of DNA (Okazaki fragments) that will be joined together to form a finished lagging strand. The experiments that led to the discovery of Okazaki fragments gave evidence for which of the following ideas? • A) DNA polymerase is a directional enzyme that synthesizes leading and lagging strands during replication. • B) DNA is a polymer consisting of four monomers: adenine, thymine, guanine, and cytosine. • C) DNA is the genetic material. • D) Bacterial replication is fundamentally different from eukaryotic replication. The key shouldn’t be way longer than the distractors.

• A) DNA polymerase is a directional enzyme that synthesizes leading and lagging strands during replication.