BIOL 101 1nd Edition Lecture 18 Outline of Last Lecture I. The Flow of Genetic InformationII. Act I: ReplicationIII. Act II: Transcription and TranslationOutline of Current Lecture I. Practice Exam II. Chapter 8III. Chapter 9IV. Chapter 10Current Lecture____1. A mouse sperm of genotype a B C D E fertilizes an egg of genotype a b c D e. What will the genotype of the zygote be? (Assume no nondisjunction). A) a/a; b/b; C/c; d/d; E/eB) a/a; B/b; C/c; D/d; E/eC) a/a; B/b; C/c; D/D; E/e____2. The chart below shows the results of different matings between jimsonweed plants thathad either purple or white flowers and spiny or smooth pods. Determine the dominant allele for the two traits.Purple Spiny White SpinyPurple SmoothWhite Smoothpurple spiny x purple spiny 94 32 28 11purple spiny x purple smooth 40 0 38 0purple spiny x white spiny 34 30 0 0purple smooth x purple smooth 0 0 36 11These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.white spiny x white spiny 0 45 0 16A) Purple and SmoothB) White and SmoothC) Purple and SpinyD) White and Spiny____3. Which of these changes in a DNA template causes an amino acid substitution in the protein in which Trp becomes Cys?A) TAC  TATB) ACC  ACAC) ACA  ACTD) AGG TGG____4. In giraffes, 2n = 62. How many tetrads (pairs of homologous chromosomes) would we observe under the microscope if we examined a cell in metaphase of meiosis I?A) 31B) 62C) 124D) Impossible to determine____5. A mutation occurs in a somatic cell allowing it to proceed through an important checkpoint in the cell cycle. Which of the following is a potential consequence of this mutation?A) Uncontrolled cell divisionB) A malignant tumorC) CancerD) A benign tumorE) All of the above____6. In the pedigree below, is this a dominant or arecessive disorder? A) RecessiveB) DominantAABB?Use the second pedigree below to answer Questions 7-8 about this RARE recessive disorder. (Assume individuals marrying into the family are NOT carriers).____7. What is the probability thatB is a carrier? A) 0B) 1/2C) 1/4D) 1/8E) 1____8. What is the probability that ifindividuals A and B have a child, they will have the geneticdisorder represented in this pedigree?F) 0G) 1/2H) 1/4I) 1/12J) 1/8____9. Looking at the cell to the right, this is a cell in_______________ and thediploid number for this cell is ___________. A) Metaphase of mitosis, 5B) Metaphase I of meiosis, 20C) Metaphase II of meiosis, 10D) Metaphase I of meiosis, 5E) None of the above____10. A molecular biologist is studying a geneat the level of the DNA that contains 30% adenine and 20% cytosine. What else can he assume about this gene?A) It contains 30% thymine and 20% guanineB) It contains 20% uracile and 30% guanineC) It contains 20% thymine and 30% guanineD) Its impossible to determine the percentage of each of the base pairs____11. True or False? Skin color is like pea pod color in that it is an example of a trait that is determined by a single gene.A) TrueB) False____12. Meiosis automatically produces diversity because ___ occurs at random.A) the number of chromosomes each daughter cell receivesB) the number of cell divisions each daughter cell undergoesAABB?C) which daughter cell receives maternal and which daughter cell receives paternal chromosomesD) which of the four daughter cells receive a complete set of genetic materialUse the picture below of cells in mitosis to answer Questions 13-14. NOTE- each cell (A-D) is in a different stage of mitosis.____13. Which cell is in metaphase of mitosis?A) AB) BC) CD) D____14. Which cell is in anaphase of mitosis?ABDA) AB) BC) CD) D____15. In the lab, scientists use a technique called PCR (polymerase chain reaction) to amplify/replicate DNA. If a scientist used PCR to replicate the double stranded molecule shown below, which two new fragments would he get?ATCCGAACCGCGTAGGCTTGGCGCA) GTCCGAACCGCG, TAGGCTTGGCGCB) ATCCGAACCGCG, TAGGCTTGGCGCC) CTCCGAACCGCG, ATGCCTTGGCGCD) ATCCGAACCGCG, ATGCCTTGGCGC____16. We would expect that an 18 nucleotide sequence ending with a stop codon would produce a polypeptide with ____ amino acids.A) 3B) 4C) 5D) 6E) 18____17. Which of the following takes place during translation?A) The conversion of genetic information from the language of nucleic acids to the languageof proteinsB) The conversion of genetic information from DNA nucleotides into RNA nucleotidesC) The conversion of genetic information from the language of proteins to the language of enzymesD) DNA replication____18. A Colbertberry plant is diploid. Each of its cells has twenty chromsomes (2n = 20). A Colbertberry plant cell enters mitosis without S phase occurring in interphase. Assuming mitosis completes, the result would be:A) Two daughter cells, each with some part of the 20 chromosomesB) One big cell with 20 chromosomesC) One big cell with 40 chromosomesD) Two small daughter cells with half the amount of cytoplasm and organelles as normal.In fruit flies, very dark (ebony) body color is determined by the e allele. The e+ allele produces normal, honey colored bodies. Heterozygotes for the two alleles have a marking called trident on their thorax but are otherwise honey colored. The e+ allele is incompletely dominant.____19. When female e+e+are crossed to males e+e flies, what is the probability that progeny will have the trident marking? A) 0B) 25%C) 50%D) 75%E) 1____20. If flies with trident marking mate with one another and 300 progeny are obtained, how many would you expect to have trident body?A) 75B) 150C) 200D) 300E) None of the