Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th,...

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Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th , 2009 Review Session – Wednesday, Feb. 4th

Transcript of Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th,...

Page 1: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Genetics 314 – Spring, 2009

Lecture 7

Reading – Chapter 13

First Exam – Friday, February 6th, 2009

Review Session – Wednesday, Feb. 4th

Page 2: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

How can transcription be stopped?

1. Bind to the DNA a. prevent separation of the DNA

strands b. block the binding site of RNA

polymerase

2. Bind to RNA polymerasea) change the shape of the RNA

polymerase so it can not function properly

b) bind or remove the sigma factor

Page 3: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Transcriptional control occurs:

naturally: binding of proteins to DNA to block transcription is a common form of gene regulation

chemically: antibiotics are available that can bind to DNA or RNA polymerase

Page 4: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Genetic Code – Chapter 13

How is the genetic information carried on the DNA?

Sequence of bases act as a code:

DNA RNA protein

bases bases amino acids

Page 5: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

How many nucleotides equal one amino acid?

Have 4 different nucleotides and 20 different amino acids

Want the fewest number of bases per amino acid for the greatest efficiency

Page 6: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

2 bases per amino acid (42) = 16 combinations

3 bases per amino acid (43) = 64 combinations

4 bases per amino acid (44) = 256 combinations

Page 7: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

So 3 bases per amino acid gives enough combinations for 20 amino acids

A 3 base sequence codes for one amino acid and is called a codon

Page 8: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

The 3 base codons for the 20 amino acids is considered to be universal in that in general the codons code for the same amino acid regardless of the species

Page 9: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

In experiments with a 3 base codon system it was shown that the code was a non-overlapping code, meaning that the bases in one codon were not part of another codon

Page 10: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

The presence of 64 codons for 20 amino acids allows for some redundancy, meaning that an amino acid may have more than one codon coding for that amino acid

Page 11: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

The fact that the codons for an amino acid were similar lead to the proposal by Crick of the ‘wobble’ hypothesis. In this hypothesis the specificity of the code is more in the first two bases allowing for variation in pairing at the third base without changing the amino acid

Page 12: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

example:

proline - CCU

CCC

CCA

CCG

This system could increase the speed of protein synthesis and reduce errors

Page 13: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Summary - Genetic Code

- 3 bases per codon

- non-overlapping code

- some degeneracy or redundancy in the code

Page 14: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Summary - Genetic Code

- can have ‘wobble’ for amino acid specificity with a greater specificity for the first two bases of a codon

- code is almost universal, a major reason for genetic engineering

Page 15: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Conversion of the genetic information into a product requires the translation of the nucleotide sequence in DNA to an amino acid sequence in a protein

This concept of one gene one enzyme was first proposed by Beadle and Tatum (1941)

Page 16: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Products of Translationpolypeptide = protein

Classes of proteins- enzymes- receptor proteins- transport proteins- structural proteins- nucleic acid binding proteins- ribosomal proteins- storage proteins

Page 17: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Protein Structure

Structure and function of a protein is controlled by the sequence of the amino acids and the interaction of the amino acids within the polypeptide and with amino acids in other polypeptides

Page 18: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Primary structure: linear sequence of amino acids

Secondary structure: interaction of amino acids in the polypeptide in the form of hydrogen bonds that result in the folding of the polypeptide into various shapes/structures

Examples: helix pleated sheets

Page 19: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Tertiary structure: additional folding of the polypeptide by covalent bonds forming between amino acid side groups

The folding due to covalent bonds will be more permanent than those found in the secondary structure, why?

Page 20: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

cyst.

S

Scyst.

Page 21: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Quaternary structure: interaction between polypeptides

Example:

enzymes with multiple sub-units

Page 22: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Structure of the protein dictates its function

Change the amino acid sequence and you may change the structure of the protein

A change in structure can lead to reduced functionality or non-functionality

Page 23: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

So changes in the base sequence of the DNA within a gene can change the functionality of the gene product if the change results in an amino acid(s) change in a critical location of the polypeptide

Page 24: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Translation – Chapter 13

Translation - The formation of a polypeptide with the amino acid sequence directed by the nucleotide sequence of a specific RNA molecule (mRNA)

Page 25: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

There are 3 types of RNA needed for translation:

- messenger RNA (mRNA)

- transfer RNA (tRNA)

- ribosomal RNA (rRNA)

Page 26: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Translation - The formation of a polypeptide with the amino acid sequence directed by the nucleotide sequence of a specific RNA molecule (mRNA)

Page 27: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

There are 3 types of RNA needed for translation:

- messenger RNA (mRNA)

- transfer RNA (tRNA)

- ribosomal RNA (rRNA)

Page 28: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Messenger RNA

- large molecular weight (500,000 +)

- intermediate carrier of the genetic code

- relatively short-lived but will vary among genes and between prokaryotes and eukaryotes

- may be translated many times

- 2 to 10% of cellular RNA

- amount of modification required prior to translation differs between prokaryotes and eukaryotes

Page 29: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Difference in mRNA between eukaryotes and prokaryotes is the processing required after transcription.

Eukaryotes Prokaryotes

Page 30: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Processing of the eukaryotic hnRNA to produce mRNA occurs in the nucleus

Processing involves:

- removing introns

- adding a guanine ‘cap’ on the 5’ end

- adding a poly-adenine tail to the 3’ end

Page 31: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.
Page 32: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

Eukaryotic gene

exon intron exon intron exon

exon - region that codes for part of the gene product

intron- region that does not code for part of the gene product

Page 33: Genetics 314 – Spring, 2009 Lecture 7 Reading – Chapter 13 First Exam – Friday, February 6 th, 2009 Review Session – Wednesday, Feb. 4th.

How the presence of introns was detected:

expected observed

mRNA mRNA

ssDNAssDNA