Solved by a verified expert:BIOL 214 Genes and EvolutionExam 2Fall 2013October 13, 2013Name:Test No: BPeriod: 2PMPlease put your multiple choice answers on the scantron sheet. Fill in your name,the Test No. as B and the period as 2PM.Note, there are short answer questions beginning on page 9 , write your answersto these questions on the exam.Your NameLetterTime1 Multiple Choice Questions, 2 points each, put your answers on the scantron.1. An operon isA. A molecule that can turn genes on and offB. An inducer bound to a repressorC. A series of regulatory sequences controlling transcription of protein-codinggenesD. A promoter, an operator, and a group of structural genesE. Any long sequence of DNA2. Which statement about selective gene transcription in eukaryotes is not true?A.B.C.D.E.Regulatory proteins can bind at a site on DNA distant from the promoter.Transcription requires transcription factors.Genes are usually transcribed as groups called operons.Both positive and negative regulation occur.Many proteins bind at the promoter.3. In prokaryotes, the concentration of cAMPA.B.C.D.E.decreases with increased glucose concentration.increases with increased glucose concentration.negatively regulates operons such as the lac operon.makes CAP.degrades CAP.4. In the E. coli lac operon, what would happen if there is a mutation in the operator?A.B.C.D.E.RNA polymerase could not bind to the DNAThe structural genes would not be expressedThe structural genes would always be expressedAllolactose could not bind to the repressor proteinThis answer is wrong and is just taking up space.5. In the E. coli trp operon,what would happen if there is a mutation in the repressorprotein gene resulting in a repressor protein that can not bind to DNA?A. The structural genes would not be expressedB. The structural genes would always be expressedC. The structural genes would only be expressed when tryptophan wasabsentD. The structural genes would only be expressed when tryptophan waspresentE. Once again, this answer is wrong and is just taking up space.2 6. In my favorite bacterial organism, an operon exists that has as its structural genesthe enzymes necessary to synthesize the amino acid histidine. Which of thefollowing conditions would result in transcription of the His operon structuralgenes?A.B.C.D.E.the presence of histidine and absence of glucosethe presence of histidinethe absence of histidinethe absence of tryptophanthe absence of both histidine and glucose7. Which of the flowing statements about enhancers is correct?A.B.C.D.E.They contain a unique base sequence called the TATA box.They are located just 5’ to the promoter.They are found in exons.They are found quite distant from the genes they control.They are part of the 3’ UTR.8. Which statement accurately describes the relationship between RNA interference,small interfering RNA (siRNA) and micro RNAs (miRNA)?A. They all describe different methods of post-translational gene regulation.B. Micro-RNAs can be broken down into two types: RNA interference moleculesand small interfering RNAs.C. RNA interference occurs in two ways: by micro RNA or by small interferingRNAs.D. Small interfering RNAs bind to dicer protein, micro RNAs do not; both are typesof RNA interference.E. Micro RNAs are transcribed from viruses, small interfering RNAs are transcribedfrom nuclear DNA; both are types of RNA interference9. Methylation regulates transcriptionA.B.C.D.E.via the addition ofvia the addition ofvia the addition ofvia the addition ofvia the addition ofmethyl groups to thymine bases of RNAa methyl group to thymine bases of DNAmethyl groups to cytosine bases of DNAmethyl groups to cysteine bases of RNAmethyl groups to cysteine in RNA polymerase3 10. What types of proteins bind to the promoter-proximal elements?A.B.C.D.E.Basal transcription factors and RNA polymeraseNegative regulators of transcription like repressorsHistone remodeling proteinsRegulatory transcription factors such as activatorsRegulatory transcription factors such as enhancers11. Micro RNAs (miRNAs)A.B.C.D.E.are short sequences only 22 or so bases long.are noncoding RNA sequences that bind to a protein complex.can regulate mRNA translation by cleaving the mRNA.can prevent translation of its target mRNA by blocking the ribosome.all of these12. How does active CAP induce expression of the genes of the lactose operon?A.B.C.D.E.It terminates production of repressor molecules.It degrades the substrate allolactose.It stimulates splicing of the encoded genes.It stimulates the binding of RNA polymerase to the promoter.It binds steroid hormones and controls translation.13. If inflated pod type is dominant in pea plants, a cross between a true breedinginflated pod plant with a true breeding constricted pod plant will result inA.B.C.D.half the plants with inflated pods, half the plants with constricted podsall inflated pods in the F1 generationall constricted pods in the F1 generationall pod shape that is intermediate in shape, slightly inflated, but with someconstrictionsE. mostly inflated pods in the F1 generation with a few constricted pods14. A sexually reproducing animal has two unlinked genes, one for foot size (L) and onefor tail length (T). It’s genotype is LlTt. Which of the following genotypes is apossible gamete from this organism?A.B.C.D.E.LLlLlTtLtTt4 15. Inheritance of the M and N blood type is best described by which of the followingtermsA.B.C.D.E.codominancemultiple allelespleiotrophyepistasisincomplete dominance16. In the example of the Labrador dogs, an epistatic gene blocks coat color and mustbe in the homozygous condition for epistasis to result. Black coat color is dominantto brown coat color. What then is the ratio of Black to Brown to Yellow (blonde)labs in a dihybrid cross?A.B.C.D.E.1:2:19: 3 : 3 : 19:4:39:3:412 : 3 : 117. If nondisjunction occurs in meiosis II during gametogenesis, what will be the resultat the completion of meiosis?A.B.C.D.E.All the gametes will be diploid.Half of the gametes will be n + 1, and half will be n – 1.one of the gametes will be n + 1, one will be n – 1, and two will be n.There will be three extra gametes.Two of the four gametes will be haploid, and two will be diploid.18. Which of the following is true of a species that has a chromosome number of2n = 16?A.B.C.D.E.The species is diploid with 32 chromosomes per cell.The species has 16 sets of chromosomes per cell.At prophase I there will be 32 separate chromosomes.Each cell has 8 homologous pairs.A gamete from this species has 4 chromosomes.5 19. After telophase I of meiosis, the chromosomal makeup of each daughter cell isA.B.C.D.E.haploid, and the chromosomes are each composed of two chromatids.haploid, and the chromosomes are each composed of a single chromatid.diploid, and the chromosomes are each composed of a single chromatid.diploid, and the chromosomes are each composed of two chromatids.tetraploid, and the chromosomes are each composed of two chromatids.20. During meiosis an error can occur resulting in gametes with incorrect chromosomenumbers. What is this type of error called?A.B.C.D.E.translocationnondisjunctionduplicationreplicationthe Philadelphia chromosome21. The following is a map of four genes on a chromosome:Between which two genes would you expect the highest frequency ofrecombination?A. A and WB. W and EC. E and GD. A and EE. A and G22. Referring to the map in the previous question, the map units are shown betweeneach pair of genes. What recombination frequency would be experimentallyobserved between W and G?A.B.C.D.E.15 %A value slightly larger than 15%A value slightly smaller than 15%30%8%6 23. In your fly lab you perform a cross between a red eyed, long winged female and apurple eyed, dumpy winged male. Which of the following offspring would representa recombinant phenotype?A.B.C.D.E.Red eyed, long winged of either sexRed eyed, long winged malePurple eyed, long winged of either sexPurple eyed, dumpy winged femaleWhite eyed, long winged of either sex24. In humans, sex-linked disorders are more common in males. Which of the followingcan account for this observation?A. Most sex-linked disorders are found on the Y chromosomeB. Males inherit one copy of the X chromosome so recessive disorders aremore likely to be phenotypically expressedC. Males inherit their X chromosome from their dad, so are likely toinherit sex-linked disorders.D. Male to male transmission is rare.E. Recessive disorders skip generations.25. Vermillion eyes is a sex-linked recessive characteristic in fruit flies. If a femalehaving vermilion eyes is crossed with a wild-type male, what percentage of the F1males will have vermillion eyes?A.B.C.D.E.0%25%50%75%100%26. In your role as a genetic counselor, you meet with a couple that is concerned abouthaving a child. Both potential parents are of normal height. The women has a sisterwith Achondroplasia (a type of dwarfisim) and the man’s mother is also anAchondroplastic dwarf. They know that dwarfisim is caused by an autosomaldominant gene and would like to know their probability of having a child withdwarfism.A.B.C.D.E.0%25%50%75%100%7 Use the following information for the next two questions:Stem length, pod color and seed shape were three characters that Mendel studied. Eachis controlled by an independently assorting gene and has a dominant and recessiveexpression as follows:CharacterStem lengthPod colorSeed shapeDominantTall (T)Green (G)Round (R)Recessivedwarf (t)yellow (g)wrinkled (r)27. If a plant that is heterozygous for all three characters is allowed to self-fertilize,what is the probability of producing progeny which are homozygous for all threetraits? (a dwarf plant with yellow pods and wrinkled seeds)A.B.C.D.E.1/91/161/643/81/3228. Again, if a plant is heterozygous for all three characters is allowed to self-fertilize,what is the probability of a plant that is tall with yellow pods and wrinkled seeds?A.B.C.D.E.1/93/323/641029. The Philadelphia chromosome is formed when there is a translocation involvingchromosomes 9 and 22. This is highly associated with a blood cancer, why is thisthe case?A. The translocation results in over activity of the ABL gene due to its position nearthe BCR gene.B. The translocation results in no activity of the ABL gene due to its position distantfrom the BCR gene.C. The translocation is not associated with the cancer cell growthD. I can’t think of any other choices, so don’t choose this one.E. This choice is just taking up space, so is incorrect.8 30. In pea plants, flowers can be either purple or white. Mendel’s experimentsshowed that blending traits was not the method of inheritance for flower colorin these plants. Which of Mendel’s observations was important to disproving theblending model of inheritance?A. The F1 plants all had pink flowersB. The F2 generation produces some plants with purple flowers and some withwhite flowersC. The number of purple to white flowering plants was equal in the F1 generationD. The number of purple to white flowers was a 9: 3: 3: 1 ratio in the F2E. This experiment did not disprove the blending hypothesis9 Short Answer Questions, write your answers on these pages.1. Radish flowers may be red, purple or white. A cross between a red-flowered plantand a white-flowered plant yields all-purple offspring. The part of the radish we eatmay be oval or long, with long being the dominant characteristic.A. Describe the phenotypes F1 progeny from a cross between a true-breeding redlong radish and a true breeding white oval radish. (4 pts)B. Determine the phenotypes and the phenotypic ratios from the F2 progenyresulting from a self-cross of the F1 progeny from part A. (4 pts)C. The flower color trait observed in these radishes is an example of what type ofinheritance? (2 pts)2. Three babies were recently mixed up in a hospital. After consideration of the databelow, assign the correct baby to each couple and write the genotypes of each baby.(8 pts)Couple #Blood groupsIA and AIIA and BIIIB and OBaby #Blood groups1B2O3AB10 3. A wild-type fruit fly (heterozygous for gray body color and normal wings) is matedwith a black fly with vestigial wings. The offspring, have the following phenotypicdistribution: wild type, 778; black body-vestigial wings, 785; black body –normalwings, 158; gray body –vestigial wings, 162.A. Using symbols consistent with the method of labeling genes in drosophila,list the genotype of the black fly with vestigial wings. (4 pts)B. Calculate the recombination frequency between body color and wing type.Show your work. Indicate the parental phenotypes and the recombinantphenotypes. (10 pts)4. Individuals affected by a condition known as polydactyly have extra fingers or toes.The following pedigree shows the pattern of inheritance fo this trait in one family:From the pedigree, can you tell if polydactyly comes from a dominant or recessiveallele? (2 pts) Is the trait sex-linked? (2 pts) What is the genotype of the individualindicated by the arrow? ( 4 pts)11
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