(8) 1. How can protease and DNase enzymes be used to determine the DNA content of nucleosome core particles? Comment on both enzymes and what they act on.
(8) 2. Describe the role of telomerase in telomere formation.
(4) 3. Trisomy 13 is known as Patau syndrome. If in a child with Patau syndrome two of the three copies of chromosome 13 are identical, when during the meiotic process did the nondisjunction event occur?
(10) 4. The following pair of chromosomes is present in an inversion heterozygote. X marks the placement of the centromere.
X---ABCDEFG X---abfedcg
If a crossover event takes place between D and E in meiosis, what will the resulting products look like? Illustrate each by sequence and describe the chromosome.
(8) 5. Contrast the functional differences between an oncogene and a tumor-suppressor gene.
(4) 6. Some XY individuals phenotypically present as females and some XX individuals as males. What chromosomal abnormality could account for this?
(4) 7. Describe tautomerism and the way in which this chemical event may lead to mutation.
(6) 8. Compare and contrast the mutagenic effects of:
b) alkylating agents and
c) base analogs.
(8) 9. Is the binding of a eukaryotic transcription factor to its DNA recognition sequence necessary and sufficient for an initiation of transcription at a regulated gene? What else plays a role in this process?
(8) 10. Explain briefly the yeast two-hybrid method and how it relates to transcriptional regulation.
(8) 11. Progression through the cell cycle depends on the interaction of two types of regulatory proteins: Kinases and cyclins. List the functions of each and describe how they interact with each other to cause cells to move through the cell cycle.
(8) 12. Describe the key experiments in the discovery of
a) cell division cycle genes
b)cell cycle checkpoint genes.
(4) 13. What metabolic advantage do the middle-repetitive sequences of rRNA and histone genes impart to the cell?
(12) 14. You have isolated a new disease causing organism, and have extracted the genetic material. The organism contained nucleic acid with a remarkable base composition. A=21%, C=29%, G=29% U=21%. When heated it showed a major hyperchromic effect, and when kinetics were studied the nucleic acid of the organism exhibited the Cot curve shown below (The complexity of T4 DNA is 105 base pairs). Analyze this information carefully and draw all possible conclusions about the genetic material of this organism, based on the above observations. What information is missing and needed to confirm your hypothesis about the nature of this molecule.
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PROBLEMS
1. DNase treatment will degrade DNA not protected by its association with histone proteins. After DNase treatment, protease will digest the histones, leaving behind those DNA sequences bound by the histones.
2. Telomerase adds telomeric DNA to the tips of eukaryotic chromosomes. Telomerase self-primes this reaction via an internal short stretch of an RNA molecule that contains sequences complementary to the repeated DNA element. The DNA element is then synthesized by DNA polymerase.
3. The nondisjunction event occurred during meiosis II, resulting in retention of two identical sister chromatids.
4. X--- ABCDefba---X (dicentric)-two centromeres
gcdEFG (acentric)-no centromere
5. Mutations in both oncogenes and tumor-suppressor genes are associated with cancer. The products of oncogenes are thought to function in normal cells as growth factors that promote and regulate cell division. The product of tumor suppressor genes is thought to repress and regulate cell division.
6. XY females have a deletion of the portion of the Y chromosome containing the testis-determining factor (TDF). XX males posses a translocation of a portion of the Y chromosome containing the TDF onto the X chromosome.
7. Tautomerism- keto to enol shift alters base pairing properties
8. a) alters hydrogen-bonding specificity of each base
b)add to DNA bases various chemical groups that either alter their base pairing properties or cause structural distortions of the DNA molecule
c) incorporate base analog during replication, they are more prone to mispairing than normal nucleotides
9. No Binding DNA must be physically linked to transcriptional activator domain, typically an acid blob
10. Separate DNA binding from acid blob.
Key words: bipartite nature of tx activator, DNA binding, transcriptional activation, your protein, library, reporter gene, phenotypic assay.
11. Cyclins periodically regulated proteins that bind kinase subunit. Their binding is essential for kinase activity.
Kinase- the kinase subunit that depends upon cyclin binding for activity.
The complex phosphorylates substrates in a cell-cycle dependent fashion.
12. a) Isolate temperature sensitive mutations. distinguish mutants that block at any stage of the cell cycle (like mutations in ribosome biogenesis) versus mutations that block with a uniform cell cycle arrest.
b) Rad Isolate mutants that are radiation sensitive. Determine which ones arrest in one cell cycle, versus mutants that continue to divide, and die as colonies of cells. These cells live hard and die fast. They could restore the viability of this class of mutants by providing additional time in the cell cycle
Mad, Bub Isolate mutants that are sensitive to microtubule poisons, benomyl. These cells continue to divide in the presence of the drug and die also as colonies of cells.
13. The gene products (proteins) of these sequences are typically required in large amounts for optimal cellular metabolism. Thus the cell can attain high protein production via multiple copies in metabolically active cells.
14. First of all, because of the presence of uracil, (U), the molecule appears to be RNA. As A/U = G/C = 1, the molecule may be a double helix. The hyperchromic shift and reassociation kinetics support this hypothesis. The kinetic study demonstrates several things. First, the shape of the Cot curve reveals that there are two frequency classes of RNA. Further the complexity, or total length of unique sequence RNA is greater than that of either phage or E. coli. The complexity can be calculated, since a direct proportionality between Cot1/2 and number of base pairs exists. Cot1/2 value for repetitive DNA = .1, cot1/2 for unique DNA = 1000. About 40% of the genome is repetitive, and 60% is unique. If complexity of t4 DNA = 105 base pairs, then
The missing information concerns the sugars. Our model predicts that ribose rather than d-ribose should be present. If not, the organism contains a very unusual molecule as its genetic material.