Joey BryanBIOL 341-02Exam 3 Extra Credit

Pledge: I hereby declare, upon my word of honor, that I have neither given nor received any unauthorized help on this work. Initials indicate acceptance of pledge: JWB

1. On Separate Sheet.

2. Fish line A had nondisjunction in meiosis I. As shown by the figures, the eggs produced with nondisjunction in meiosis I all result in aneuploidy, which would give them an odd shape but allow them to be viable.

3. If nondisjunction occurred in mitosis, then half of the resulting cells would be normal and half would exhibit aneuploidy, which is not the case for either line A or B.

4. Deletion loss of a chromosome segment

5. Reciprocal translocation movement of genetic material between nonhomologous chromosomes in which there is a two-way exchange.

6. Robertsonian translocation long arms of two acrocentric chromosomes become joined to a common centromere through translocation creating one metacentric chromosome with two long arms and another chromosome with two very short arms (that is usually lost)

7. Pericentric inversion chromosome segment is turned 180 degrees and includes the centromere

8. At the end of a linear chromosome there is no adjacent stretch of replicated DNA to provide the essential 3-OH group, which will not allow DNA nucleotides to replace the primer. This means that each chromosome would become shorter and shorter after each round of replication and would eventually destabilize.

9. A does not refer to the problem because the end replication problem occurs in both leading and lagging strands. B refers to the problem because the problem occurs at the ends of the chromosomes. C does not refer to the problem because DNA replication is not ended after one round of replication (but rather has the potential after many rounds of replication).

10. Dispersive. This is the only method of replication where there would only be one visible band using this test. Both conservative and semiconservative would produce two visible bands.

11. A False. DNA polymerase can only add nucleotides to the 3 end of a growing DNA strand.B False. Helicase unwinds both strands at the same time and rate.C False. DNA polymerase only adds nucleotides to the 3 end.D True. It is a reason for the occurrence of the lagging strand because transcription will take place in the opposite direction of the unwinding in one strand, causing discontinuous transcription in that strand.

12. TBP binds to the TATA box to bend and partially unwind the DNA.

13. The end result of phosphorylation of the carboxy terminal tail of RNA polymerase is elongation or the RNA.

14. Letter A appropriately describes the process of 5 capping.

15. A consensus sequence of AAUAAA is upstream of the cleavage site and determines the point where cleave will take place. A large number of proteins find the cleavage site and remove the 3 end. After cleavage, adenine nucleotides are added to the 3 end, creating the poly(A) tail. The correct answer is d.

16. A True. Transcription for these genes starts in the center and moves outward. Because transcription only takes place from 5 to 3, the transcription for each gene must take place on opposite strands.B False. Genes are only transcribed from 5 to 3.C False. Genes are only transcribed from 5 to 3.D False. Because transcription for these genes takes place from the center and moves outward and transcription only moves from 5 to 3, this transcription for each gene must take place on different strands.E False. Genes are only transcribed from 5 to 3.

17. snRNA. Some convert pre-mRNA to mRNA.

18. T site

19. Aminoacyl (A) site receives all tRNAs (except the initiator tRNA).Peptidyl (P) site At this site, adjacent tRNAs amino acids form a dipeptide bond and the amino acid on the tRNA at this site is released. The tRNA from the A site slides into the P site.Exit (E) site from this site the tRNA moves into the cytoplasm where it can be recharged with another amino acid

20. In Eukaryotes, translation is initiated by an initiation complex (including ribosome, initiation factors, and initiator tRNA with its amino acid) that binds to the 5 cap. It then scans the mRNA for the first AUG codon.

21. Genes function by encoding enzymes, and each gene encodes a separate enzyme. 1) Some genes encode for structural proteins and not enzymes. 2) Some proteins are composed of more than one polypeptide chain, and different genes encode for these different chains.

22. All sequences in the DNA that are transcribed into RNA.

23. A consensus sequence is a sequence in a promoter that possesses considerable similarity with other sequences in the promoter. It represents the most commonly encountered nucleotides at each of these positions. These sequences are important in the initiation of transcription.

24. NCARG/CA/TGCTG/TNCAYCGTNCC

25. Core promoter = 15-17. Consensus and near transcription site.Regulatory promoter = 7-9. Consensus and further upstream of transcription site than core promoter.

26. There would be no change. Base A and C both occur at the same rate at position 11, so transcription would not be altered.

27. Rate of transcription would increase because RNA polymerase is more likely to bind to C than it is to T at position 5 because C is more prevalent.

28. Yes, transcription would not be changed because A and T occur at the same percentages.

29. DNA polymerase I removes and replaces primers while DNA polymerase III elongates the DNA.

30. The lagging strand has Okazaki fragments that are linked together to create a continuous new DNA molecule while the leading strand does not.

Extra, Extra Credit

1) A. I did not realize that trisomy was a form of aneuploidy because I did not fully understand the definition of aneuploidy. I could picture the figures for nondisjunction that are located in the book; I just couldnt relate them to the specific example given.

2) H. I could not remember during which phase of transcription 5 capping occurred. All choices were almost identical making process of elimination tough.

3) I. The wording of the question and given choices c and d made this question difficult since I did not fully understand the process how a poly(A) tail is synthesized.

4) M. This one caused my problems because I did not study initiation of translation well enough.

5) O. This was the single-most confusing question on the test. I had no idea how to interpret the bar graph to come up with the consensus sequence. I am still not sure exactly how to do so.