MCDB 4650 Developmental Control of Gene Expression

1. MCDB 4650Developmental Control of Gene Expression

2. A reliable way to compare the genomic information content between two different kinds of organisms would be to: a) Compare the total amounts of DNA in the two haploid genomes. b) Compare the complexity (number of nucleotides in non-repetitive DNA sequences) in the two genomes. c) Sequence the two genomes, count the number of putative genes based on computer analysis, etc., and compare these numbers. d) Count the number of different proteins that are synthesized throughout each animal's life cycle and compare these numbers. e) None of the above.

3. The pufferfish is, obviously, a fish, and a member of the vertebrate class of animals, as we are. The complexity of the human genome is about 3,000 megabases (Mb), while that of the pufferfish is about 400 Mb. Which two of the following reasons are most likely to contribute significantly to this difference? Briefly explain your choices. a) Pufferfish have many fewer genes than humans. b) Pufferfish genes have smaller and fewer introns than the corresponding human genes. c) Pufferfish genes have smaller and fewer exons than the corresponding human genes. d) Pufferfish genes have smaller and fewer regulatory sequences than human genes. e) Pufferfish have less intergenic DNA than humans.

4. 1 2 3 The gene diagrammed below is expressed in two different cell types, A and B. In cell type A, the resulting protein includes the sequences encoded by all three exons, but in cell type B, the protein contains only the sequences encoded by exons 1 and 3. The splicing that occurs during mRNA processing in B cells must be: i. from donor sequence of exon 1 to acceptor sequence of exon 2 ii. from acceptor sequence of exon 1 to donor sequence of exon 3 iii. from donor sequence of exon 1 to acceptor sequence of exon 3 iv. from donor sequence of exon 2 to acceptor sequence of exon 3. v. from acceptor sequence of exon 2 to donor sequence of exon 3.

5. How do early embryonic divisions differ from cell divisions that normally take place in the soma? a) there are no differences b) embryonic divisions occur faster c) the early embryonic divisions are not accompanied by cell growth d) all of the above e) none of the above

6. Two different cell types may contain different amounts of the same protein because they differ in: a) rates of transcription initiation for the corresponding gene. b) alternative splicing of the corresponding transcript to give a different 3’ UTR. c) alternative splicing of the corresponding transcript to give a different 5’ UTR. d) a, b and c e) a and b f) b and c g) a and c

7. Transcriptionally active chromatin is: a) compacted, because DNA looping is required to activate transcription complexes. b) relaxed, because transcription factors must be able to gain access to DNA sequences.

8. Acetylation of histones in the vicinity of a gene causes the gene to be a) activated because the histones bind less tightly to the DNA than unacetylated histones. b) silenced because the histones bind less tightly to the DNA than unacetylated histones. c) activated because the histones bind more tightly to the DNA than unacetylated histones. d) silenced because the histones bind more tightly to the DNA than unacetylated histones.

9. Acetylation of histones in the vicinity of a gene causes activation of the gene by decreasing histone binding to DNA, because acetyl groups linked to amino acid side chains a) sterically hinder the binding. b) eliminate positively charged carboxyl groups. c) eliminate positively charged amino groups. d) eliminate negatively charged carboxyl groups. e) eliminate negatively charged amino groups.

10. a) an amino group b) a carboxyl group c) an acetyl group

11. a) an amino group b) a carboxyl group c) an acetyl group

12. a) an amino group b) a carboxyl group c) an acetyl group

13. Transcription factors generally control the rate of: a) transcription initiation. b) transcript elongation. c) transcription termination. d) transcript stability. e) transcript processing.

14. Do all exons code for protein? a) Yes b) No

15. Cellular oncogenes, in which dominant gain-of-function mutations can cause tumor formation, encode proteins whose normal function is to a) inhibit cell division or prevent apoptosis. b) promote cell division or prevent apoptosis. c) inhibit cell division or initiate apoptosis. d) promote cell division or initiate apoptosis.

16. True or false? The fairly common observation that the same transcription factor can activate transcription of a gene in one cell type and repress transcription of the same gene in another cell type can be explained by: 1) the presence of different DNA regulatory (response) elements for that gene in the two different cell types. 2) different combinations of other transcription factors present in the two cell types. a) Both these statements are true. b) 1 is true; 2 is false. c) 1 is false; 2 is true. d) Both are false.