EXAMPLE QUESTIONS AND ANSWERS1. Topoisomerase does which one of the following?(a) Makes new DNA strands.(b) Unties knots in DNA molecules.(c) Joins the ends of double-stranded DNA molecules.(d) Is required for making recombinant DNA molecules in vitro.2. Which one of the following general properties of the genetic code makes it impossiblefor sequence information to be transferred from protein to RNA or DNA?(a) It is a triplet code.(b) It is commaless.(c) It is degenerate.(d) It includes start and stop codons.3. The following enzymes or enzyme activities are involved in DNA replicationand repair:A. DNA polymeraseB. endonucleaseC. exonucleaseD. ligaseE. helicaseF. single-strand binding proteinG. topoisomerase(a) Which enzyme(s) make 5' - 3' phosphodiester bonds? A, D, G (b) Which enzyme(s) untie knots in DNA molecules? G (c) Which enzyme(s) make single-strand breaks in DNA backbones? B, G (d) Which enzyme(s) join the ends of double-stranded DNA molecules? D 4. The drawing below represents a replicating DNA molecules from an animal seen in theelectron microscope. Which number(s) represent the origin(s) of replication? 3, 4 5. Draw one or more arrows to show the direction(s) in which sequence information canNOT flow.DNARNA ProteinDNARNAProtein6. What is the N-terminal amino acid in β-globin? Methionine The N-terminal amino acid is the first amino acid with which translation begins.Translation of all mRNAs begins at the 5’ end of the mRNA with the AUG start codonwhich codes for Met.7. Below is a diagram of a segment of DNA and its RNA products. Identify some of theregions by writing the letter corresponding to the name of each in the parentheses.a exon b intron c 3’ UTR d 5' flanking region e operon f 3' flanking regiong 5’ UTR h transcription start i ATG j TAG k transcription stop8. In the blank nest to the name of each kind of mutation on the left, write the number(s)from the list on the right that tell what effect the mutation will most likely have on apolypeptide.(a) missense 3 1 polypeptide is longer2 polypeptide is shorter(b) nonsense 2 3 polypeptide has one amino acid substitution4 polypeptide has many amino acid substitutions(c) frameshift 4, 1, 2 9. Recall that in peas, G and g determine yellow and green peas, respectively, and T and tdetermine tall and short plants. The genes are unlinked. A plant of genotype G g T t iscrossed to a short plant that developed from green seeds.(a) Write the genotype of the second parent. g g t t (b) Out of 100 progeny seeds, how many do you expect to be yellow and producetall plants?G g T t X g g t t --> 1/4 G g T t 1/4 G g t t 1/4 g g T t 1/4 g g t t1/4 X 100 = 2510. Martian dragons are normally green, but a recessive mutation makes them white. Thealleles are w+ and w. This mutation is due to a 500-bp insertion. Suppose you took a fewcells from the tail of a sleeping green Martian dragon (very gently!), then isolated DNAfrom the cells, used PCR to amplify the region containing the insertion, and separated thePCR fragments on an agarose gel by electrophoresis. You got two bands, one of size 400bp and the other 900 bp.(a) Circle the genotype of this dragon: w+ w+ w+ w w w (b) If this green dragon mated with a white dragon, what would be the genotypes andgenotypic ratios in the offspring? (Assume no sex linkage.)w+ w X w w --> 1/2 w+ w 1/2 w w 11. A three-factor test cross was done between a Drosophila female heterozygous forblack body (b), purple eye (pr), and engrailed (en) and a homozygous recessive male. Thenumbers and phenotypes of the progeny were:+ + + 431 Pb en pr 437 P+ + pr 1 D+ en + 37b en + 2 D+ en pr28b + pr 35b + + 29 1000(a) Write P next to each parental genotype.and D next to each double crossover type.(b) Which gene is in the middle? Pr (c) What is the map distance between the en and pr loci? 7.5 map units or cM7 ptsShow your calculations. 1 75/1000 = 0.07537 2357512. In Drosophila, the e (for ebony) gene is on chromosome III. On what chromosomewould you find the e+ allele? Circle the correct number:I (X) II III IV13. Below are partial sequences (50 bp) of the Adh gene from three differentindividuals of Drosophila melanogaster (Dm 1, 2, 3) and one copy from Drosophilavirilis (Dv). (a) Calculate the sequence difference between each of the following pairs ofgenes:Dm 1 and Dm 2 4/50 = 0.08 Dm1 and Dm 3 1/50 = 0.02 Dm2 and Dm 3 3/50 = 0.06 (b) Calculate the nucleotide diversity in D. melanogaster.(0.08 + 0.02 + 0.06)/3 = 0.0533or(4 + 1 + 3)/3 X 50 = 0.053314. In the preceding question, the three copies of the Adh gene from D.melanogaster are more similar to each other than they are to the D. virilis gene. Thebest explanation is:(a) The D. melanogaster sequences have a more recent common ancestor.(b) The mutation rate is lower in D. melanogaster(c) The population size is larger in D.melanogaster.(d) Balancing selection is operating in both species.THE (ALMOST) UNIVERSAL GENETIC CODE U C A G UUU phe F UCU ser S UAU tyr Y UGU cys C UU UUC phe F UCC ser S UAC tyr Y UGC cys C CUUA leu L UCA ser S UAA och ZUGA opa ZAUUG leu L UCG ser S UAG amb ZUGG trp W G CUU leu L CCU pro P CAU his H CGU arg R UC CUC leu L CCC pro P CAC his H CGC arg R CCUA leu L CCA pro P CAA gln Q CGA arg R ACUG leu L CCG pro P CAG gln Q CGG arg R G AUU ile I ACU thr T AAU asn N AGU ser S UA AUC ile I ACC thr T AAC asn N AGC arg R CAUA ile I ACA thr T AAA lys K AGA arg R AAUG met M ACG thr T AAG lys K AGG arg R G GUU val V GCU ala A GAU asp D GGU gly G UG GUC val V GCC ala A GAC asp D GGC gly G CGUA val V GCA ala A GAA glu E GGA gly G AGUG val V GCG ala A GAG glu E GGG gly G