Unit 1 Principles of Molecular Biology & Gene Expression PDF

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Summary

Explain genetic recombination and genetic
mapping.
• Review the steps involved in transcription.
• Review the process of translation and
protein production....

Description

APPLIED MOLECULAR BIOLOGY

Principles of Molecular Biology & Gene Expression

BIOT2027

Unit 1

Unit 1 Outcomes

Genetic Recombination

• Explain genetic recombination and genetic mapping. • Review the steps involved in transcription. • Review the process of translation and protein production.

• Recombination: crossing over between two homologous chromosomes. – Produces new combinations of alleles – Involves the cutting and covalent joining of DNA sequences

• Occurs in meiosis I • The offspring have a combination of alleles not seen in the parents and are termed recombinants.

Genetic Recombination

Genetic Recombination

• All DNA is recombinant DNA

Homologous recombination is an essential cellular process catalyzed by enzymes specifically made and regulated for this purpose.

Homologous recombination is conserved across all three domains of life, as well as viruses, suggesting that it is a nearly universal biological mechanism.

Recombination provides:

• Genetic exchange works constantly to blend and rearrange alleles, involving the physical exchange of DNA sequences between chromosomes. Recombination crossovers are responsible for genetic variation.

• Genetic variation • Replacing or repairing damaged DNA • Restarting stalled or damaged replication forks • Regulation of expression of some genes

© 2014 Oxford University Press

Genetic Mapping

Genetic Recombination

• Thomas Hunt Morgan suggested that genes are arranged on a chromosome in a linear fashion, and so genes that are further apart have a higher likelihood of recombining. • This proved to be true and mathematical analysis of the frequency of recombination allows for the determination of the distance between genes on a chromosome.

• The frequency of crossing over between two genes on the same chromosome depends on the physical distance between these genes. Long distances gives the highest frequencies of exchange. Therefore, the farther apart two genes are on a chromosome, the more likely they are to recombine © 2014 Oxford University Press

Genetic Mapping • In this way chromosomes can be mapped to show the location of each gene on every chromosome. – Unit of distance in genetic maps = centimorgans, cM – 1 cM = 1% chance of recombination between markers

Genetic Mapping A genetic map assigns DNA fragments to chromosomes Determines the linear order of genes on the chromosome

Chromosome Maps: http://www.ncbi.nlm.nih.gov/books/NBK22266/ https://www.ncbi.nlm.nih.gov/projects/genome/gui de/human/index.shtml

Genetic Mapping • The distance between genes on a genetic map are measured in map units Map units are related to the frequency of recombination between the genes

• If the frequency of recombination between two genes is found to be 5%, the genes are said to be separated by 5 map units

Create a Genetic Map Example Given the crossover frequency of each of the genes on the chart, construct a map of the arrangement of the genes on a chromosome. Genes

Frequency of recombination

AC

30

BC

45

BD

40

AD

25

Create a Genetic Map Problem If six genes (A-F) gave the following frequencies of recombination, map the arrangement of the genes on a chromosome. Genes

Frequency of recombination

AB

1.5

AC

2.5

AD

0.2

AE

3.5

AF

2.0

BC

1.0

CD

2.3

BE

2.0

EF

5.5

The Central Dogma

What do Genes Do? 1. Information storage (DNA à RNA à Protein) 2. Replicate (when cells divide) 3. Mutate (the basis of evolution)

Nucleic Acids • Responsible for storing and transmitting genetic information • Two types: – Deoxyribonucleic acid (DNA) – Ribonucleic acid (RNA)

Chromosomes carry all of the genetic information in the form of DNA, which is transcribed into RNA, which is translated into protein

Transcription

Gene Expression 2 major steps involved: • Transcription = Synthesis of messenger RNA (mRNA) that is complementary to one of the strands of DNA (coding strand) representing the gene

Three key stages: 1. Initiation 2. Elongation 3. Termination

• Translation = Assembly of a protein (polypeptide) based on information encoded in the mRNA

Step 1: Initiation of Transcription • RNA polymerase recognizes a region upstream of the gene called the promoter • RNA polymerase binds the promoter tightly, causing localized separation (~12 bp) of the DNA strand à transcription bubble

RNA polymerase

Step 1: Initiation of Transcription • Next, the RNA chain is built using ribonucleotide triphosphates (ATP, GTP, CTP, UTP). • After the first nucleotide is in place (usually a purine), the RNA polymerase adds a second and so on • Several nucleotides of the new RNA chain may be added together before the RNA polymerase leaves the promoter

Initiation in Prokaryotes • Single RNA polymerase • Polymerase requires an initiation factor at the promoter site: sigma (s)

Initiation in Eukaryotes • RNA polymerase I, II, & III • RNA Pol II à transcribes all protein-encoding genes into mRNA Promoter Structure Two parts: • Core promoter à TATA box (–25) • Upstream promoter elements

Initiation in Prokaryotes Promoter Structure

Sigma

• Two conserved sequences of 6-7 bp in length upstream of transcription start site • –35 box • –10 box

Initiation in Eukaryotes • RNA Pol requires several general transcription factors (GTFs) to initiate transcription • RNA Pol + 6 GTFs = preinitiation complex – Forms at TATA box – 6 GTFs: TFIIA, TFIIB, TFIIC, TFIID, TFIIE, TFIIF, TFIIH

• Help RNA Pol bind to the promoter and melt (open) DNA

Step 2: Elongation • RNA polymerase moves along the DNA template locally separating the DNA template, adding to the growing RNA strand in the 5’ à 3’ direction. • As RNA polymerase passes, the two DNA strands anneal. • RNA polymerase has proof reading functions.

Step 3: Termination • The region at the end of the gene is called a terminator. • Essentially, opposite of promoter • RNA polymerase, RNA, and DNA dissociate

• Elongation occurs with very similar mechanisms in prokaryotes and eukaryotes.

Termination in Prokaryotes Two Types: 1. (r) Rho-Independent (intrinsic) – A short inverted repeat (20 nucleotides) results in a stem-loop structure – Stem-loop interacts with RNAP causing it to pause – Followed by a stretch of adenine nucleotides, which are weakly bound and cause the RNA Polymerase to dissociate from RNA and DNA

Termination in Prokaryotes 2. (r) Rho-dependent – Ring shaped protein – Binds to mRNA and follows along behind RNA Pol – When hairpin forms and DNA pauses, rho acts as a helicase and causes the DNA and RNA to unwind and dissociate, releasing the mRNA – When RNAP pauses because of the hairpin, Rho is able to ‘catch up’ with the RNAP and dissociate the RNA from the DNA

Termination in Eukaryotes •

•

•

Termination protein – RNAP I (rRNA) – A termination protein binds to the DNA at a specific site and physically blocks RNAP I from further transcription – RNAP I, rRNA, and DNA dissociate from each other and transcription stops “Torpedo model” – RNAP II (mRNA) – RNAP continues past termination site, but endonucleolytic cleavage occurs in poly(A) site – Second (downstream) RNA strand is uncapped – Recognized by an RNase which quickly degrades remaining RNA

Termination in Eukaryotes Torpedo model

Polyadenylation (Allosteric) model

Polyadenylation of 3’ end – RNAP III (tRNA) – Lesser known model, linked to termination of transcription – Series of weak A=U binding causes dissociation (no stem-loop/pausing required)

Summary of Transcription

RNA Processing in Eukaryotes 1. Capping of the 5’ end – Addition of methylated guanine base to 5’ end of RNA – Occurs immediately after initiation

2. Splicing 3. Polyadenylation of 3’ end – Addition of many adenine residues to 3’ end – Termination of transcription

RNA Splicing in Eurkaryotes Eukaryotic genes are mosaics • Exons à coding sequences • Introns à noncoding sequences RNA splicing: Pre-mRNA à Mature Pre-mRNA can be spliced in more than one way = more than one polypeptide product

Translation – Overview

RNA polymerase

DNA

Translation

Protein Transcription

Interpretation of genetic information contained in mRNA to direct the synthesis of a protein • Highly conserved across all organisms • Involves: 1. Ribosomes 2. Transfer RNA (tRNA) 3. mRNA

Ribosomes

Ribosomes Large subunit

• Protein synthesizing machines – E. coli à 2 subunits make up ribosome à 50S & 30S – Eukaryotes à 2 subunits make up ribosome à 60S & 40S – “S” is a sedimentation coefficient

• 50S + 30S = 70S (not 80S because S not related to weight)

• Peptidyl transferase center – formation of peptide bonds

Small subunit • Decoding center

Three binding sites for tRNA 30S

30S 50S

50S

Transfer RNA (tRNA) • Translates the nucleic acid language (mRNA) translate into protein language • tRNA is the “bifunctional molecule” • tRNA recognizes both mRNA and amino acids • “Cloverleaf model” à schematic representation

1. Amino-acyl (A) site 2. Peptidyl (P) site 3. Exit (E) site

Amino acid attachment at 3’ OH group Acceptor stem

tRNA anticodon pairs to complementary mRNA codon

Transfer RNA (tRNA) • 3’ tail – attachment point for amino acids • Catalyzed by group of enzymes named aminoacyl-tRNA synthetases (there are 20, one for each amino acid)

Codons • Codons are tri-nucleotide sequences of mRNA • The “reading frame” is the grouping, or order, in which the three-nucleotides are read and translated into amino acids • Every triplet of nucleotides specifies an amino acid • “Start” and “Stop” codons signify the beginning and end of amino acid (polypeptide) chain

Transfer RNA (tRNA) • At the “bottom” of tRNA, a 3 base pair sequence of nucleotides pair with complementary 3 base pair in mRNA (anticodon loop) • The 3 mRNA nucleotides are “codons” and the complement tRNA nucleotides are “anticodons” • For example, mRNA codon for amino acid alanine (Ala) attaches to anticodon on tRNA. This tells the ribosome to add alanine.

Codons For example, the following mRNA sequence illustrates a start and stop codon, reading frame (read 5’ à 3’) and translated protein of 4 amino acids in length Start codon

Stop codon

5’ – AAUGAAAUUGAAUUAAAUU – 3’ Met - Lys - Leu - Asn

mRNA translation

Note: Stop codon does not code for an amino acid like the start codon does Each sequence could potentially be read in 3 reading frames

Codon Chart

Practice Question Use the codon chart to determine the polypeptide sequence that would be produced from this mRNA strand:

CGAUAUGGAACGCAUCCCGAGACUGAUAUAAGC

Initiating Protein Synthesis

Practice Question Use the codon chart to determine the polypeptide sequence that would be produced from this mRNA strand: Start codon

Stop codon

CGAUAUGGAACGCAUCCCGAGACUGAUAUAAGC Met – Glu – Arg – Ile – Pro – Arg – Leu – Ile Methionine – Glutamate – Arginine – Isoleucine – Proline – Arginine – Leucine – Isoleucine

Three events must occur: 1. Ribosome recruited to mRNA 2. Charged tRNA must be placed in P site 3. Ribosome must be precisely positioned over start codon (sets reading frame)

Initiating Protein Synthesis – Prokaryotes

Initiating Protein Synthesis – Eukaryotes

5’ of start codon à ribosome binding site (RBS)

• 5’ methylated guanine cap recruits small ribosome subunit – tRNA is already bound • Bound ribosome “scans” for AUG start codon • Once bound to start codon, the large subunit binds • Poly-A tail enhances the level of translation – promotes efficient recycling of ribosomes

• Also called the “Shine-Delgarno sequence”

Initiator tRNA (fMet-tRNA) – located at P site base pairs with AUG start codon (AUG = Methionine) • fMet-tRNA is an N-formylated methionine

Prokaryotic

Eukaryotic

rRNA from the small subunit of the ribosome recognizes RBS sequence on the mRNA, resulting in mRNA being recruited to the ribosome to begin translation.

5’ methylated cap recruits the mRNA to the ribosome. The tRNA with the UAC anticodon for MET is already bound to the ribosome, recognizes the mRNA start codon, and begins translation.

Translation Elongation Ribosome cycle: 1. Aminoacyl-tRNA binding 2. Peptide bond formation 3. Translocation

Translation Elongation (3 steps) • Initially, the amino acid-tRNA (fMet/Met-tRNA) is bound to the P site on ribosome • Elongation begins when a second AA-tRNA binds to the A site on the ribosome. • Peptide bond formed between first AA (fMet/Met) and the amino acid part of AA-tRNA in the A site. This is done by the enzyme peptidyl transferase. Result: dipeptidyl-tRNA in A site • Finally, mRNA is moved 1 codon length through ribosome (translocated). In doing so, dipeptidyltRNA is moved from A to P site. The empty tRNA that used to contain fMet/Met is moved to E site, then it leaves ribosome.

Termination of Translation 3 stop codons (UAG, UAA, and UGA) • Recognized by release factors (RFs) à activate hydrolysis of polypeptide from pepidyl-tRNA

Summary of Protein Translation

Image References: • • • • •

https://www.mun.ca/biology/desmid/brian/BIOL2060/BIOL2060-20/CB20.html http://www.nature.com/scitable/topicpage/each-organism-s-traits-are-inherited-from-6524917 http://bioweb.wku.edu/courses/biol114/vfly1.asp https://www.biologycorner.com/worksheets/genetic_maps.html http://karimedalla.wordpress.com/2013/02/21/10-3-gene-linkage-polygenic-inheritance/

All Oxford University Press images are from: Craig, N.L., Cohen-Fix, O., Green, R., Greider, C., Storz, G., & Wolberger, C. (2014). Molecular Biology: Principle of Genome Function. Oxford University Press.

Transcription and Translation...