BIOL 225 Genetics-Final Exam November 17, 2006 Dr. Sandra Davis INSTRUCTIONS: 1. Read the questions carefully and write your answers in the space provided. If you need more space, clearly indicate WHERE the rest of the answer is located (for example, on the back of the same page). If there is something that you do not wish me to count, (for example, if you make an error) please cross it out. 2. Read each question carefully before starting to answer it so you don t overlook any additional instructions. If you get stuck on a question, go on to another question and return to the original question later. It is a good strategy to read over the entire exam and then select the questions you feel most confident about to answer first. 3. In your answers to problems that require you to calculate a numerical answer, you must show how you set up your calculation to receive full credit for your final numerical solution. 4. A blank sheet of paper has been provided for you at the end of the exam which you may use as scratch paper. GOOD LUCK! Question #1: _ (20 pts.) Question #2: _ (16 pts.) Question #3: _ (12 pts.) Question #4: _ (12 pts.) Question #5: _ (15 pts.) Question #6: _ (12 pts.) Question #7: _ (13 pts.) Bonus: _ (5 pts.) TOTAL: _ (100 pts.) ------------------------------------------------------------- NAME: _ Please print legibly ID NUMBER: _ SIGNATURE: _
1. For each of the following, choose the one alternative that best completes the statement or answers the question (2 pts. each, 30 pts total). _ A mutation changes a codon from UCG (ser) to UAG (stop). This is a _ mutation. A.) point B.) nonsense C.) silent D.) both A and B E.) all of the above _ Where would you find the Pribnow Box? A.) in 5 UTR region of a prokaryotic mrna B.) in 3 UTR region of a prokaryotic mrna C.) in the terminator region of a eukaryotic gene D.) in the promoter region of a prokaryotic gene E.) in the promoter region of a eukaryotic gene An allopolyploid A.) is the result of unequal crossing over B.) has chromosomes from at least 2 different species C.) has multiple chromosomes from the same species D.) is always sterile E.) is always inviable Which of the following is true of duplications? A.) they increase the amount of genetic material B.) they can lead to genes with novel functions C.) they interfere with crossing over in heterozygotes D.) B and C E.) A and C _ Which of the following is true of translation? A.) Initiation involves a trna bind to the start codon in the P site of the ribosome. B.) New amino acids are added to the growing polypeptide chain when trnas bind to their appropriate codon in the A site of the ribosome C.) It takes place in the cytoplasm of both prokaryotes and eukaryotes D.) mrna is read in the 5 3 direction E.) all of the above The mutagen 5-bromo-uracil is a(n) A.) base-modifying agent B.) intercalating agent C.) base-analog D.) deaminating agent E.) free radical Which of the following is true about a Robertsonian translocation? A.) they decrease the amount of genetic material in a carrier B.) they can lead to genes with novel functions C.) a carrier may be phenotypically normal D.) B and C E.) A and C
_ In eukaryotic chromosomes, DNA is wrapped around a single core that consists of A.) RNA B.) histone proteins C.) collagen proteins D.) polysaccharides E.) heterochromatin _ In Drosophila, males flies that are hemizygous for the Bar allele (genotype BY) A.) are phenotypically normal B.) lack a functional copy of the Bar gene on their X chromosome C.) have two copies of the Bar gene on their X chromosome D.) have a pericentric inversion within the Bar gene E.) none of the above _ During mismatch repair of a double-stranded DNA molecule, the first step is A.) an exonuclease removes nucleotides from the unmethylated strand of DNA B.) an exonuclease removes nucleotides from the methylated strand of DNA C.) DNA polymerase removes the mismatched nucleotide from the 3 end of the mismatched strand D.) the enzyme photolyase is activated by exposure to UV light E.) helicase unwinds the DNA near the mismatched region
2. (A ) An RNA molecule has the following percentages of bases: A=23 %, U=42 %, C=21 %, G=14%. i.) Is this RNA single stranded or double stranded? How can you tell? (2 pts.) ii.) What would be the percentages of bases in the template strand of DNA that contains the gene for this RNA? (0.5 pts. each, 2 pts. total) A = T = _ G = _ C = _ (B. ) The following nucleotide sequence is found on the template strand of DNA: Sequence of DNA template: 3'--TAC TGG CCG TTA GTT GAT ATA ACT--5' nucleotide number------------>1 24 i.) When RNA polymerase transcribed this DNA strand did it move from the left to the right side of this molecule or from the right to left side of this molecule? How do you know? (2 pts.) ii.) What will be the sequence of the mrna molecule transcribed from this DNA molecule? (2 pts.) iii.) Determine the amino acids of the protein encoded by this sequence, using the table of the genetic code provided at the end of this exam (2 pts.).
2. cont iv.) Give the altered amino acid sequence of the protein that will be found in each of the following mutations (1 pt. each, 4 pts. total). a. Mutant 1: A transition at nucleotide 11. b. Mutant 2: A transition at nucleotide 13. c. Mutant 3: A one nucleotide deletion at nucleotide 7. d. Mutant 5: A transition at nucleotide 9. v.) Write the letter of the above mutations (a d) that correspond to the following types of mutations (0.5 pts. each, 2 pts. total): frameshift silent missense nonsense
3. This question concerns a specific type of structural rearrangement occurring in chromosomes in humans. These are the chromosomes with which you are going to illustrate pairing, and for which you are going to diagram the gametes produced at the end of meiosis. (a ) Suppose an individual has acquired a translocation between autosomes #14 and #21, as illustrated in the diagram above. What is the name of this type of chromosomal abnormality? Be careful- it may not be what you first assume!! (1 pt.) (b ) Show how the chromosomes would potentially line up in Metaphase I of meiosis. Be sure that all chromosomes, chromatids, and chromosome parts are clearly labeled. (3 pts.)
3. cont (c ) Diagram of gametes produced by this individual, including all chromosomes, chromosome parts, chromatids, clearly labeled. (4 pts.) (d ) If the individual who possessed the translocated chromosomes illustrated above mated with an individual who possessed a completely normal set of chromosomes show the combinations of zygotes produced during fertilization and indicate whether or not the offspring would be viable or what type of chromosomal abnormality it might have (1 pt. per diagram; 4 pts. total).
4. Identify the structural change present in each of the chromosomes below, being as specific as possible with your answers. The first diagram shows the ORIGINAL chromosome, before any changes occurred. (2 pts. each; 12 pts. total)
5. Mr and Mrs Lambert have not yet been able to produce a viable child. They have had two miscarriages and one severely defective child that died soon after birth. Studies of the banded chromosomes of father, mother and child showed that all chromosomes were normal except for pair number 6. The number 6 chromosomes of the family are shown in the following figure: (a ) Which parent shows a chromosomal abnormality? (1 pt.) (b ) What is the name of this type of chromosomal abnormality (be specific!)? (2 pts.) (c ) Why has this abnormality gone undetected until this individual stared trying to have children? (3 pts.) (d ) From which parent did the longer chromosome 6 of the child come from? (1 pt.) (e ) From which parent did the shorter chromosome 6 of the child come from? (1 pt.) (f ) Since neither parent has a chromosome like the shorter chromosome of the child, how did this chromosome arise? Be specific as to what events in the parent gave rise to this chromosome (3 pts.). (g ) Why is the child not phenotypically normal? (2 pts.) (h) For the parent with the chromosomal abnormality, what can be predicted about his or her potential to have a healthy offspring? (2 pts.)
6. Red-green colorblindness is an X-linked recessive disorder in humans. A young man with Klinefelter syndrome is colorblind. Both of his parents have normal colorvision. (a ) Give the genotypes with respect to colorvision of each of the individuals described above (1 pt. each, 3 pts. total): Man with Klinefelter syndrome: Mother: Father: (b ) In which parent did the nondisjunction occur to give rise to the man with Klinefelter syndrome? Explain how you came to this conclusion (3 pts.). (c ) When during gamete formation did the nondisjunction take place? Explain how you came to this conclusion (3 pts.). (d ) In humans, how many chromosomes would you find in the nuclei of cells that are (0.5 pts. each, 3 pts. total): Monoploid: Triploid: Nullisomic: Doubly monosomic: Tetrasomic: Carriers of a Robertsonian translocation:
7. In the plant genus Triticum, there are many different polyploid species, as well as diploid species. Crosses can be made between the various species to produce hybrids. Three examples of these crosses are shown below, along with the number of chromosomes found in the resulting hybrid offspring: Cross Number of chromosomes in the hybrid offspring T. turgidum X T. monococcum 21 T. aestivum X T. monococcum 21 T. turgidum X T. aestivum 14 (a ) How many chromosomes would you find in the following? (1 pts. each, 4 pts. total) A gamete of T. monococcum? A somatic cell of T. aestivum? A gamete of T. turgidum? A somatic cell of T. monococcum? (b ) The hybrid from the T. turgidum X T. aestivum cross has an even number of chromosomes (14) but attempts at mating it show that it is sterile. Explain (3 pts.). (c ) For T. monococcum (1 pt. each, 2 pts. total): x = _ n = _ (d ) In fact, all three of the hybrids shown in the table are sterile. Explain the steps you would go through to create a fertile hybrid from T. turgidum and T. monococcum. At each step be sure to indicate the number and type of chromosomes present in intermediate steps and in the final fertile hybrid (4 pts.).
Bonus: Six bands in a salivary gland chromosome (chromosome number 2) of Drosophila are shown in the following figure, along with the extent of 5 deletions (Del 1 to Del 5) found in different flies. Recessive alleles a, b, c, d, e, and f are known to be in the region of these deletions, but their order is not known. When crosses are made to produce heterozygotes for the 5 deletions and each of the different recessive alleles (i.e., a deletion on one of the chromosome number 2 s and the recessive allele on the other chromosome number 2), the following results are obtained: a b c d e f Del 1 - - - + + + Del 2 - + - + + + Del 3 - + - + - + Del 4 + + - - - + Del 5 + + + - - - In this table, a minus sign means that the deletion is missing the corresponding wild-type allele and the heterozygous fly shows the recessive phenotype (the deletion uncovers the recessive), and a plus sign means that the corresponding wild-type allele is still present on the chromosome number 2 with the deletion. Determine the order of the genes on the chromosome and the band on the chromosome that contains each gene: Order of the genes (2 pts): Bands associated with each gene (0.5 pts. Each, 3 pts. Total): Band 1 2 3 4 5 6 Gene
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