FSU PCB 3063 - Genetics Study Guide: Exam 3

Unformatted text preview:

Genetics Study Guide Exam 3 Genetic Code Transcription Chapter 12 1 Transcription process by which base sequence in DNA is converted into RNA The enzyme responsible is RNA polymerase 2 The genetic code was deciphered using synthetic mRNAs and in vitro translation in vitro cell free protein synthesizing system and an enzyme used to understand the code polynucleotide phosphorylase enabled production of synthetic mRNAs Nirenberg and matthei added RNA homopolymers to the in vitro translation system to decipher which amino acids were encoded by the first few codons based on which amino acids were incorporated into the polypeptide these mRNAs are templates for polypeptide synthesis in the cell free system In cell free system amino acids are incorporated into polypeptide chains must contain essential factors for protein synthesis in the cell ribosomes tRNA amino acid and other molecules essential to translation to trace protein synthesis the amino acid must be radioactive finally an mRNA must be added serves as template that will be translated Repeating Copolymers technique used to decipher the genetic code allows determination of specific sequences of triplet codons designating specific amino acids In vitro translation of the repeating copolymer CUCUCUCUCUCU gives leucine and serine CCUCCUCCUCCUCCU Gives leucine serine prolin It is already known that CUU codes for leucine What are the codons for serine and proline Serine Proline A ucc ucu ccu B ccu ucc ucu C cuc ccu CUCUCUCUCUCU gives leucine and serine codons CUC and UCU CCUCCUCCUCCUCCU Gives leucine serine proline codons CCU CUC and UCC It is already known that CUU codes for leucine Because CUU leu then CUC leu So UCU must ser and therefore UCC ser Thus CCU pro 3 Base Transition mutations in which a purine is switched with another purine A with G Or A pyrimidine is switched with another pyrimidine U with C Base Transversion mutations in which a purine is switched with a pyrimidine or vice versa Transitions are five time more likely than transversions 4 Genetic Code written in linear form The genetic code is written in linear form using the ribonucleotide bases that compose mRNA molecules as letters The ribonucleotide sequence is derived from the complementary nucleotide bases in DNA Each word within the mRNA consists of three ribonucleotide letters referred to a triplet code With several exceptions each group of three ribonucleotides called a codon specifies one amino acid The code is unambiguous each triplet specifies only a single amino acid The code is degenerate that is a given amino acid can be specified by more than one triplet codon This is the case for 18 of the 20 amino acids The code contains one start and three stop signals triplets that initiate and terminate translation No internal punctuation such as a comma is used in the code Thus the code is said to be commaless Once translation of mRNA begins the codons are read one after the other with no breaks between them The code is nonoverlapping Once translation commences any single ribonucleotide at a specific location within the mRNA is part of only one triplet The code is nearly universal With only minor exceptions almost all viruses prokaryotes archea and eukaryotes use a single coding dictionary i ii iii iv v vi vii viii ix Insertions or deletions of one or two nucleotides resulted in frameshift mutations Insertion or deletion of three nucleotides resulted in insertion or deletion of a single amino acid and did not shift the reading frame explains that genetic code is in triplets There are 64 amino acids but only 61 code for actual codons while the other three are stop codons Degenerate an amino acid can be specified by more than one triplet codon 18 of the 20 amino acids Example Both the codons UUU and UUC specify the amino acid phenylalanine Universal code is nearly universal with only minor exceptions Viruses prokaryotes archae and eukaryotes all existing life has a common ancestor The code is degenerate and ordered which help protect against point mutations Most mutations base transitions affecting the 3rd position do not change the amino acid Mutations base transitions affecting the second position often result in replacing one amino acid with a similar one Base substitution mutations can be missense changes the amino acid nonsense changes an amino acid codon into a stop codon silent no amino acid change 5 Start codon AUG codes for methionine as the first amino acid in a polypeptide chain N formyl methionine fmet initiator codon rarely GUG can specify met during initiation 6 RNA polymerase requirements to be a polymerase enzyme 5 to 3 synthesis DNA template no primer needed contains ribose Substrate same as DNA polymerase EXCEPT substrate nucleotides contain ribose rather than deoxyribose form of the sugar 7 Promoter when RNA polymerase subunit recognizes specific DNA sequences located in the region upstream 5 from point of initial transcription from gene Bacterial promoters contain two sequences 10 TATA and the 35 sequence that RNA polymerase binds to initiate transcription It explores a length of DNA until it recognizes the promoter region and binds to about 60 nucleotide pairs of the helix Once this occurs the helix is denatured unwound making the DNA template accessible to the action of the enzyme It is a base sequence at the front of a gene that RNA polymerase binds to in order to begin transcription A specific nucleotide sequence in DNA that binds in RNA polymerase positioning it to start transcribing RNA at the appropriate place 8 Three steps in processing eukaryotic mRNA s and function 1 Capping To protect the transcript from degradation and it provides a site for ribosome binding in the cytoplasm Protects 5 end of molecule from nuclease attack Also involved in transport of mature mRNAs across the nuclear membrane into the cytoplasm and in the intiation of the translation into the mRNA into protein 2 Addition of poly A tail The 5 cap and polyA tail are necessary for mRNA stability and transport of the nucleus In it s absence the RNA transcripts are quickly degraded 3 Removal of introns splicing Introns intervening noncoding sequenced within the gene Exons the coding sequences Splicing introns are cut out and exons are joined Capping and addition of poly A tail are necessary for mRNA stability and transport out of the nucleus 9 Introns intervening noncoding sequences within the gene Exons the coding sequences Splicing Introns cut out and exons joined


View Full Document

FSU PCB 3063 - Genetics Study Guide: Exam 3

Documents in this Course
Meiosis

Meiosis

22 pages

Meiosis

Meiosis

35 pages

Genetics

Genetics

11 pages

Exam 3

Exam 3

13 pages

Exam 4

Exam 4

24 pages

Exam #3

Exam #3

30 pages

Genetics

Genetics

22 pages

Exam 3

Exam 3

12 pages

Exam 3

Exam 3

12 pages

Exam 3

Exam 3

12 pages

Exam 3

Exam 3

12 pages

Exam 3

Exam 3

12 pages

Exam 2

Exam 2

16 pages

Exam 2

Exam 2

16 pages

Exam 2

Exam 2

16 pages

Exam 2

Exam 2

16 pages

Exam 2

Exam 2

16 pages

Exam 1

Exam 1

19 pages

Genetics

Genetics

24 pages

Chapter 1

Chapter 1

21 pages

Chapter 1

Chapter 1

20 pages

Genetics

Genetics

50 pages

Genetics

Genetics

20 pages

Exam 3

Exam 3

12 pages

Exam 1

Exam 1

19 pages

Exam 1

Exam 1

19 pages

Exam 1

Exam 1

10 pages

Exam 3

Exam 3

30 pages

Chapter 9

Chapter 9

30 pages

Exam 2

Exam 2

16 pages

Exam 1

Exam 1

46 pages

Exam 3

Exam 3

12 pages

Load more
Download Genetics Study Guide: Exam 3
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Genetics Study Guide: Exam 3 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Genetics Study Guide: Exam 3 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?