Exam 2 Biology II Winter 2013 Name: Notice that on the last page of this exam, you will find a genetic code table and some potentially useful sequences. You can tear it off to make it more useful. This page may also be used as scratch paper if you like. Multiple Choice Questions. Circle the one best answer for each question unless otherwise noted. (2 points each) 1. Many people think cystic fibrosis (CF) results from inheriting a CF gene. This is a misunderstanding because: A. CF actually results from inheriting a non-functional protein. B. CF is not a genetic disease: it results from insufficient secretion. C. CF results from the inheritance of a specific allele. D. CF results from a mutation, not from the inheritance of a gene. 2. The best illustration of Mendel s Law of Segregation is: A. An individual with genotype Aa produces ½ A gametes and ½ a gametes. B. Mating of two Aa individuals gives a ¼ chance of an aa offspring. C. An individual with genotype AaBb produces AB, Ab, ab, and ab gametes. D. An Aa individual would show the phenotype of the dominant A allele. E. An individual with two co-dominant alleles (e.g., A L and A M ) shows both phenotypes. 3. The best illustration of Mendel s Law of Independent Assortment is: A. An individual with genotype Aa produces ½ A gametes and ½ a gametes. B. Mating of two Aa individuals gives a ¼ chance of an aa offspring. C. An individual with genotype AaBb produces AB, Ab, ab, and ab gametes. D. An Aa individual would show the phenotype of the dominant A allele. E. An individual with two co-dominant alleles (e.g., A L and A M ) shows both phenotypes. 4. An enzyme that would not be found in the nucleus of a eukaryotic cell is: A. DNA ligase B. RNA polymerase II C. RNA polymerase I D. topoisomerase (also known as DNA gyrase) E. none of the above: all of these would be found in the eukaryotic nucleus. 5. A dominant allele is: A. the most common allele of a particular gene. B. the normal allele. C. the allele that determines the phenotype of a heterozygote. D. the allele that encodes the non-functional form of a protein. Version A p. 1 of 7 February 13, 2013
6. In order to start transcription of an mrna, eukaryotic RNA polymerase looks for: A. the -10 and -35 sequences B. the Shine-Dalgarno sequence C. the TATA box D. transcription factors E. the start codon 7. If we sequenced the DNA from cells in many different tissues of the human body, we would find that: A. The CFTR gene is found only in epithelial cells of the lung, pancreas and other secretory tissues. B. Every cell has one copy of the CFTR gene. C. Every cell has two copies of the CFTR gene. D. The CFTR gene is in the nucleus of most cells but in the cytoplasm of lung and pancreas cells. 8. Which of the following is capable of breaking hydrogen bonds between two DNA strands? Choose all that apply. A. DNA polymerase III B. helicase C. RNA polymerase D. primase E. initiator protein 9. For a particular gene, which of these would be the longest molecule? A. mrna in the cytoplasm B. mrna in the nucleus C. coding sequence D. exon 10. Which of the following does not occur during RNA processing in eukaryotes? A. addition of a poly(a) tail B. removal of introns C. removal of RNA primers D. addition of a G nucleotide to the 5 phosphate 11. Which investigators showed that one gene encodes one enzyme? A. Watson and Crick B. Beadle and Tatum C. Hershey and Chase D. Meselson and Stahl 12. Avery, McCarty and MacLeod s experiments provided evidence that DNA is the genetic material by: A. showing that Streptococcus pneumoniae was not transformed by DNAse treated cell extract. B. showing that heat-treated S cells could still transform R cells. C. showing that radioactive phosphorus did not enter virus-infected bacteria. D. showing that radioactive sulfur did not enter virus-infected bacteria. E. showing that DNA replication is semiconservative. Version A p. 2 of 7 February 13, 2013
13. Put the following terms in size order, from largest to smallest (3 points): protein, chromosome, virus, nucleotide, eukaryotic cell, mitochondrion, phospholipid 14. From the list below, choose four DNA or RNA sequences and put them in the left column of blanks below. Then, choose the proteins that bind these sequences and match them up by putting them in the appropriate blanks in the right column (5 points). origin enhancer sequences proteins helicase sigma factor 30S subunit histone terminator start codon Shine-Dalgarno TBP -10 and -35 intron transcription factor initiator primase TATA box 15. Bob and his wife Jane like chocolate. Bob s mother is a real chocoholic: she loves, loves, loves chocolate (and has no real interest in seeking treatment). Bob s father, though, never cared for chocolate. Bob s first child (a son) and second child (a daughter) like chocolate, and son #3 is a chocoholic like his grandmother, but to Bob s surprise, daughter #4 doesn t like chocolate. a. Assuming taste for chocolate is a genetic trait, draw a pedigree for this family in the space at right (3 points). b. Determine dominance and give specific evidence to support your claim (2 points). c. Define appropriate genetic symbols (2 points). d. Give everyone in the pedigree a genotype (2 points). 16. In the appropriate boxes in the replication fork diagram at right: a. Label the leading and lagging strands (2 points). b. Label the indicated end (3 or 5 ) (1 point). c. Name the indicated enzyme (1 point). Version A p. 3 of 7 February 13, 2013
17. You are studying six human genes, which we can symbolize simply with letters A-E. a. If a female has the genotype AABbccDdEE, what are the gametes that she can make? (2 points) b. Suppose she marries a male who has hairy feet, a trait resulting from the d allele, and is heterozygous for the other five genes. What is the probability that they will have a child with hairy feet? (2 points) 18. Primary ciliary dyskinesia (PCD) is a disease in which ciliated cells normally involved in protecting the respiratory tract and in hearing fail to function, resulting in deafness and susceptibility to lung infections. It is caused by mutations in the DNAI1 gene on chromosome 9. Below is a segment of DNAI1, starting with the +1 nucleotide, and the same region from a PCD patient. Normal: PCD: TCTTCAGACGAGGGAGCGTTTTGTAGGCTCTCCAGGGGTTGAGATGATTCCTGCTTCTGC TCTTCAGACGAGGGAGCGTTTTGTATGCTCTCCAGGGGTTGAGATGATTCCTGCTTCTGC a. What do we call one gene with multiple effects (e.g., respiratory and hearing defects)? (1 point) b. What mutation has occurred in the PCD patient s DNA? (1 point) c. How would we classify this mutation in terms of its effect on the DNA? (1 point) d. How would we classify this mutation in terms of its effect on the protein? (2 points) e. Would you expect the PCD patient s allele to be a dominant or recessive allele? Explain. (2 points) 19. The sequence below is the first part of an mrna from a bacterial cell, starting with the +1 nucleotide: UGAGCAUCCCAUGGACCAUGAUAGGAGGUCCACAUGGUCCA a. There are three potential start codons. Circle the one that will actually be used and tell how you know that this is the correct one. (2 points) b. Suppose this gene were introduced into the nucleus of a eukaryotic cell and correctly transcribed to produce this same mrna. Would the same protein be produced? Why or why not? (2 points) Version A p. 4 of 7 February 13, 2013
20. The bright red color of Chippy the Cardinal s feathers is controlled by a gene, R, that has three alleles: R R produces red pigment, R B produces blue pigment, and r produces no pigment (leading to white feathers). Chippy s mother was a blue cardinal, and Chippy s father was the red son of a white mother. a. What is Chippy s genotype? (1 point) b. If Chippy has 19 brothers and sisters, what would you predict their genotypes and phenotypes would be? (4 points) 21. In cats, black fur is dominant over brown. But, there s a second gene which determines the intensity of the color. Cats homozygous for the recessive allele of this second gene have a dilute phenotype, either grey or tan. You have a grey male cat whose mother was brown and a brown female whose mother was tan. Since you re feeding them anyway, you decide they should contribute to the household income, so you breed them and sell the kittens. You know that grey or tan kittens are the most popular, so you price these at $25 each, with browns selling for $10 and the less-popular blacks priced at $5. If you were able to obtain 40 kittens (from several litters), how much income would you expect? (5 points) 22. A woman with type O+ blood has a child with type A blood. She is not sure which of two men is the father, so she consults a genetic counselor who suggests that the two men be blood-typed as an initial step in the investigation. Mr. X has type B+ blood, while Mr. Y has type AB+ blood. a. Which of these men can be ruled out, and why? (2 points) b. What is the genotype of the other man? (2 points) Version A p. 5 of 7 February 13, 2013
23. It may surprise you to know that your dislike of Brussel sprouts has a biochemical basis! The gene for the enzyme glucosinolate isomerase (GluIso) has 10 exons that make up the CDS for this protein. If all 10 exons (especially exons 4 and 5) are present, the protein is localized to the cell membrane and highly functional: it makes Brussel sprouts taste less bitter. You have isolated molecules from your own cells to figure out why your GluIso doesn t appear to work properly, and you have found that stable GluIso mrna is made and translated into protein. However, your experiments revealed that most of the GluIso protein is not associated with the cell membrane but is secreted into the extracellular environment. a. What is one modification of the GluIso mrna that stabilizes it? (1 point) b. Where in the cell would you expect the GluIso protein to be translated? (1 point) c. Briefly describe a mechanism involving RNA processing that could have led to the phenotype described for your mutant allele (you may use illustrations to support your answer). (3 points) 24. For the three RNA polymerases listed below, describe what type of RNA is produced. (1 point each) Homo sapiens RNA Polymerase III: Escherichia coli RNA Polymerase: Yeast RNA Polymerase I: 25. Gene expression for any given gene in Prokaryotes must eventually be turned off. Describe in detail how bacteria can turn off gene expression by terminating (you may use illustrations to support answers): Transcription (3 points): Translation (3 points): Version A p. 6 of 7 February 13, 2013
The genetic code: Some possibly useful sequences: 10 sequence: 5 TATAAT 35 sequence: 5 TTGACAT TATA box: 5 TATAAA Shine-Dalgarno: 5 AGGAGG Version A p. 7 of 7 February 13, 2013