Practice Test

Exam 2 Genetics, Spring 2017 1. The Central Dogma of Molecular Biology is used to describe the flow of genetic information within a cell. Draw a flow chart of the central dogma and identify the processes involved in each step. (4) DNA RNA Protein Transcription Translation 2a. A mRNA molecule has the codon 5’-GCA. What is the anticodon found on the tRNA that complements this codon? (2) 3’ – CGU – 5’ 2b. What amino acid would be attached to this tRNA? (2) Alanine 3. How does a cell identify regions of the genome that should be transcribed? (4) Upstream of all genes is a promoter region that contains binding sites that are recognized by RNA Polymerase. 4. The nitrogenous bases of RNA and DNA nucleotides fall into two chemical categories. What are those categories and list which category each base belongs in. (4) Purines = Adenine and Guanine Pyrimidines = Cytosine, Thymine, Uracil 5. Use the dsDNA sequence below to answer the following questions. 3’ AAATTCGCATTCGAATGCGGGCGGCTTAGCAATAGACGAAGGTGTAACCA 5’ 5’ TTTAAGCGTAAGCTTACGCCCGCCGAATCGTTATCTGCTTCCACATTGGT 3’ 5a. During replication, the replication fork moves through this sequence from left to right and the complement to the bottom strand is synthesized in fragments. Label the 5’ and 3’ ends of each strand. (2) 5b. This segment of DNA includes the entire coding region of a gene. Which is the template strand and which is the coding strand? (3) Bottom is template, top is coding 5c. What is the amino acid sequence of the protein encoded by this gene? (3) Met Trp Lys Gln Ile Thr Ile Arg Arg Ala 5d. Imagine there is a mutation in the 12th base from the left where instead of a C top/ G bottom pair you now have an A top / T bottom pair. How will this change the protein encoded? (3) It won’t affect the protein at all since the mutation is not within the coding region. 6. You are studying two variants of the same protein which have the following sequences: Variant 1: Met His Phe Gln His Asn Glu Tyr Trp Variant 2: Met His Ile Gln His Asn Glu Tyr Trp 6a. What is the mRNA sequence for each of the variants? For any position in which more than one nucleotide is possible, use the designation “Pu” for any nucleotide that could be either Purine, “Py” for any nucleotide that could be either Pyrimidine, and “N” for any nucleotide that could be either a Purine or a Pyrimidine. Be sure to label the 5’ and 3’ ends.(4) Variant 1: 5’ – AUG CAPy UUPy CAPu CAPy AAPy GAPu UAPy UGG – 3’ Variant 2: 5’ – AUG CAPy AUN CAPu CAPy AAPy GAPu UAPy UGG – 3’ 6b. What kind of mutation resulted in the difference between the protein variants? (2) A nonsynonymous transversion occurred in the 7th nucleotide position from the 5’ end. 7. List and describe the different types of mutations to chromosome structure. (8) Deletion – when a portion of a chromosome is missing. Duplication – when a portion of a chromosome is repeated. Inversion – when a portion of a chromosome is in the opposite orientation. Translocation – when a portion of a chromosome is on a non-homologous chromosome. Reciprocal – when 2 non-homologous chromosomes exchange parts. Non-reciprocal – when one chromosome contains a portion of a non-homologous chromosome. 8a. You insert genomic DNA from the killifish Fundulus heteroclitus into expression vectors containing a bacterial promoter and then transform yeast cells with these recombinant vectors. You use an antibody for the killifish protein Superoxide Dismutase to screen the colonies. Would this work to find the colony transformed with the gene for Superoxide Dismutase? Why or why not? (3) No. Since the plasmid has a bacterial promoter, it will not be recognized by RNA Pol. In the eukaryotic yeast cell. Therefore, the gene will not be transcribed. 8b. Would your answer change if you had transformed E. coli cells with these same recombinant vectors? Why or why not? (3) The ultimate answer is still no, but for a different reason. The bacteria will transcribe the inserted DNA in this case, but since genomic DNA was used the mRNA will contain intron sequences. Therefore, the Protein made will not be superoxide dismutatse. 9a. List two methods that can be used to isolate a specific gene and make millions of copies of it. (2) PCR Molecular Cloning 9b. List two methods that can be used to determine what traits are affected when a gene is absent from the genome or when it cannot be expressed. (2) RNAi Knockout 10. In what ways is a primary transcript modified in eukaryotic cells prior to translation? (4) A cap (inverted guanine diphosphate) is bound to the 5’ end and the cap and the first 2 nucleotides get Methylated. Then the introns are removed and the exons spliced together into a single unit. Lastly, a String of Adenines is added to the 3’ end. 11a. How does a mRNA molecule in E. coli bind in the correct location of a ribosome? (3) A sequence of nucleotides in the 5’ UTR known as the Shine-Dalgarno Sequence complements to Nucleotides of a rRNA in the small subunit. This places it in the correct location such that the start codon will be in the P-site. 11b. To which tRNA binding site on the ribosome does the initiator tRNA bind? (2) P 11c. A tRNA moves through the binding sites of the ribosome in what order?(3) APE 12. Describe how the process of excision repair works and how this process can actually make a mutation permanent as opposed to returning the DNA to the original sequence. (8) If an incorrect base is added during DNA replication, endonucleases will cut out a segment of ssDNA surrounding the mismatched base pair. A DNA Polymerase will then use the remaining ssDNA as a template to replace the nucleotides that were removed. Ligase forms the last phosphodiester bond to complete the process. This process will return a DNA molecule to the original (parental) sequence as long as the portion removed was from the newly synthesized strand since that is where the mutation had to happen. If the endonucleases cut out the ssDNA of the parent strand, then the excision repair process will change the sequence permanently since it will contain the mutation. 13. Describe the process by which a prokaryotic cell makes a copy of its chromosome. (10) DNA replication in prokaryotes starts at a fixed origin known as OriC and proceeds in both directions until hitting the terminus on the opposite side of the circular chromosome. Replication is initiated with the binding of DnaA, which breaks a few H bonds on either side of OriC. The enzyme helicase (DnaB) enters the open area on either side of DnaA and moves in both directions away from OriC breaking the H bonds between complementary nucleotides on opposite strands. This forms a replication bubble around OriC where the DNA is single stranded and each strand can be effectively replicated by the addition of complementary nucleotides. Single strand binding proteins bind to the ssDNA in the replication bubble to keep the DNA from reannealing to form dsDNA, and topoisomerases break and reform phosphodiester bonds ahead of helicase to relieve torsional stress on the DNA as it is unwound. Since DNA polymerase III can only add nucleotides to other nucleotides, the enzyme primase adds a short RNA primer to the 3’ ends of each strand on both sides of the replication bubble. A DNA polymerase attaches to each primer/DNA complex and they begin adding nucleotides that are complementary to each of the original DNA strands . These nucleotides are in the form of deoxynucleotide triphosphates (dNTPs), and DNA Pol III catalyzes the reaction of forming a phosphodiester bond between the 3’ OH group on the deoxyribose of an already bound nucleotide and the first phosphate attached to the 5’ carbon on the deoxyribose of the incoming dNTP. The outer two phosphates are cleaved from the dNTP providing the energy for the polymerase reaction. Since DNA polymerase only catalyzes this reaction in the 5’ to 3’ direction, and the DNA strands are antiparallel to each other, one strand gets replicated in a continuous fashion and the other gets replicated in fragments. The strand that is continuously replicated is called the leading strand. On the leading strand, the DNA polymerase is adding nucleotides in the same direction as the replication fork is moving. The lagging strand is replicated in fragments, since the replication fork is moving in the direction opposite of which the DNA polymerase can add nucleotides. Therefore, an RNA primer is added to the lagging strand at the replication fork and DNA polymerase moves in the opposite direction adding complementary nucleotides until it hits double stranded DNA. The polymerase then detaches from the strand and rebinds near the replication fork on a new primer. It then adds complementary nucleotides until it reaches the beginning of the previously replicated fragment. DNA Polymerase I removes the RNA primers from the newly made strand utilizing its 5’ – 3’ exonuclease activity, and replaces them with DNA. Phosphodiester bonds are then formed between adjacent fragments by the enzyme DNA ligase. Once the entire chromosome has been replicated the DNA polymerases detach. This leaves two identical chromosomes. 14. You have used PCR to amplify the locus known as Rag1 in eight different individuals. Following PCR, you performed a restriction digest on the PCR products using the restriction enzyme MluI. Below is a picture of the gel showing the results of the restriction digest. Use the information shown to answer the questions below. 14a. How many different alleles of Rag1 are found in this data set? Make a code to identify each allele. (3) 3, R1, R2 and R3 14b. Draw a restriction map for each of the alleles. (3) R1 = _____________________________ 1200 bp R2 = __________________/___________ 800 bp 400 bp R3 = _______________/______________ 700 bp 500 bp 14c. What is the genotype of each individual in this data set? (3) 1 = R1R2 2 = R2R2 3 = R1R1 4 = R1R3 5 = R3R3 6 = R2R3 7 = R1R2 8 = R3R3 Match the term on the left with the correct definition on the right. (1 each) SNP __t____ a. an enzyme that adds a sequence of adenines to the 3’ end of a mRNA molecule ß-pleated sheet ___e___ b. a DNA nucleotide that is lacking an OH on both the 2’ and 3’ carbon RNA polymerase ___d___ c. a coding region within a gene d. the enzyme that catalyzes transcription Aneuploid __l____ e. a secondary structure of a protein f. a ssDNA molecule that is labeled to allow identification of cDNA library __m____ complementary pieces of ssDNA g. the concentration of purines is equal to the concentration of snRNA _h_____ pyrimidines in a DNA molecule h. molecules that are involved in exon splicing Probe _f_____ i. organisms in which a gene from a different organism has been inserted to confer a new function ddNTP __b____ j. a ribozyme involved in binding amino acids together k. when bacterial cells ingest recombinant DNA plasmids Chargaff’s rule __g____ l. cells that contain an extra copy of one or a couple of Individual chromosomes Intron __o____ m. a collection of clones that contain recombinant DNA made by reverse transcription of mRNA n. the location on an enzyme where the substrate binds o. non-coding regions within the coding sequence of a gene p. the part of a mRNA molecule that binds to the ribosome q. an organism that contains more than two complete sets of chromosomes that originate from different species r. an organism in which both functional copies of a gene have been replaced by non-functional copies s. a short piece of DNA that provides a binding site for Taq polymerase during PCR t. a single base pair within a DNA sequence that differs among members of a population u. a spiral shaped molecule Bonus: Name three of the “final four” teams in the NCAA men’s basketball playoff (5). Get one extra point if you can name which of the four teams is your professor’s PhD alma mater. University of South Carolina = Alma mater University of North Carolina Gonzaga University University of Oregon Genetic Code U C A G U UUU - Phe UUC - Phe UUA - Leu UUG - Leu UCU - Ser UCC – Ser UCA - Ser UCG – Ser UAU – Tyr UAC – Tyr UAA – stop UAG – stop UGU – Cys UGC – Cys UGA – stop UGG – Trp C CUU – Leu CUC – Leu CUA – Leu CUG – Leu CCU – Pro CCC – Pro CCA – Pro CCG – Pro CAU – His CAC – His CAA – Gln CAG – Gln CGU - Arg CGC – Arg CGA – Arg CGG – Arg A AUU – Ile AUC – Ile AUA – Ile AUG – Met ACU – Thr ACC – Thr ACA – Thr ACG – Thr AAU – Asn AAC – Asn AAA – Lys AAG – Lys AGU – Ser AGC – Ser AGA – Arg AGG – Arg G GUU – Val GUC – Val GUA – Val GUG – Val GCU – Ala GCC – Ala GCA – Ala GCG – Ala GAU – Asp GAC – Asp GAA – Glu GAG – Glu GGU – Gly GGC – Gly GGA – Gly GGG - Gly